how to find the tension of a string

Learning Objectives

By the end of this section, you will be able to:

• Define normal and tension forces.
• Utilize Newton'south laws of motility to solve issues involving a variety of forces.
• Use trigonometric identities to resolve weight into components.

Forces are given many names, such as push, pull, thrust, lift, weight, friction, and tension. Traditionally, forces have been grouped into several categories and given names relating to their source, how they are transmitted, or their effects. The about important of these categories are discussed in this section, together with some interesting applications. Further examples of forces are discussed afterward in this text.

Normal Strength

Weight (also called forcefulness of gravity) is a pervasive force that acts at all times and must be counteracted to keep an object from falling. You definitely notice that you must back up the weight of a heavy object by pushing up on information technology when you concord it stationary, as illustrated in Effigy i(a). But how do inanimate objects like a tabular array support the weight of a mass placed on them, such as shown in Figure ane(b)? When the bag of dog food is placed on the table, the table actually sags slightly under the load. This would be noticeable if the load were placed on a card tabular array, but even rigid objects deform when a force is applied to them. Unless the object is deformed beyond its limit, it will exert a restoring force much similar a deformed spring (or trampoline or diving board). The greater the deformation, the greater the restoring force. And then when the load is placed on the tabular array, the table sags until the restoring force becomes equally big as the weight of the load. At this point the net external force on the load is cipher. That is the situation when the load is stationary on the table. The table sags quickly, and the sag is slight so we practice not detect it. Just information technology is like to the sagging of a trampoline when you climb onto it.

Figure ane. (a) The person holding the bag of dog food must supply an upwards forcefulness F_manus equal in magnitude and opposite in direction to the weight of the nutrient w. (b) The card tabular array sags when the dog nutrient is placed on information technology, much like a stiff trampoline. Elastic restoring forces in the table abound as it sags until they supply a force North equal in magnitude and opposite in direction to the weight of the load.

We must conclude that any supports a load, be it breathing or not, must supply an upwards force equal to the weight of the load, as we assumed in a few of the previous examples. If the force supporting a load is perpendicular to the surface of contact between the load and its support, this strength is defined to be a normal force and here is given the symbol Northward. (This is non the unit of measurement for force N.) The word normal means perpendicular to a surface. The normal force tin be less than the object'south weight if the object is on an incline, as yous will see in the next example.

Common Misconception: Normal Force (Northward) vs. Newton (N)

In this department we have introduced the quantity normal force, which is represented by the variable Northward. This should not be dislocated with the symbol for the newton, which is also represented by the alphabetic character N. These symbols are particularly important to distinguish because the units of a normal strength (Due north) happen to exist newtons (North). For example, the normal forcefulness N that the flooring exerts on a chair might be N = 100 N. One of import divergence is that normal force is a vector, while the newton is but a unit of measurement. Be careful not to confuse these letters in your calculations! You will encounter more similarities amongst variables and units as you go on in physics. Another example of this is the quantity work (W) and the unit watts (West).

Example 1. Weight on an Incline, a Two-Dimensional Problem

Consider the skier on a gradient shown in Figure 2. Her mass including equipment is 60.0 kg. (a) What is her acceleration if friction is negligible? (b) What is her dispatch if friction is known to be 45.0 Northward?

Effigy 2. Since motion and friction are parallel to the gradient, it is most user-friendly to project all forces onto a coordinate system where ane axis is parallel to the slope and the other is perpendicular (axes shown to left of skier). Due north is perpendicular to the slope and f is parallel to the slope, but w has components along both axes, namely due west_⊥ and [latex]\textbf{w}_{\parallel}[/latex]. Due north is equal in magnitude to w_⊥ , and then that there is no movement perpendicular to the slope, but f is less than w∥, and then that there is a downslope acceleration (along the parallel axis).

Strategy

This is a two-dimensional trouble, since the forces on the skier (the system of interest) are not parallel. The approach we have used in two-dimensional kinematics also works very well here. Cull a convenient coordinate system and project the vectors onto its axes, creating 2 connected one-dimensional problems to solve. The nigh convenient coordinate system for move on an incline is one that has one coordinate parallel to the slope and one perpendicular to the gradient. (Remember that motions along mutually perpendicular axes are contained.) We apply the symbols ⊥ and ∥ to represent perpendicular and parallel, respectively. This choice of axes simplifies this blazon of trouble, because there is no move perpendicular to the slope and considering friction is ever parallel to the surface between two objects. The just external forces acting on the system are the skier's weight, friction, and the support of the slope, respectively labeled w, f, and N in Figure two. N is always perpendicular to the slope, and f is parallel to it. Merely due west is non in the direction of either axis, then the outset step we take is to project information technology into components forth the called axes, defining westward _∥ to be the component of weight parallel to the gradient and west _⊥ the component of weight perpendicular to the gradient. Once this is done, we can consider the two carve up bug of forces parallel to the slope and forces perpendicular to the slope.

Solution

The magnitude of the component of the weight parallel to the slope is [latex]{west}_{\parallel }={west} \sin({ 25}^{\circ}) = mg\sin ({ 25}^{\circ})[/latex], and the magnitude of the component of the weight perpendicular to the slope is [latex]{w}_{\perp}={west}\cos({25}^{\circ}) = mg\cos({25}^{\circ})[/latex].

(a) Neglecting friction. Since the acceleration is parallel to the slope, we need only consider forces parallel to the slope. (Forces perpendicular to the slope add to zero, since there is no acceleration in that direction.) The forces parallel to the gradient are the amount of the skier's weight parallel to the slope westward _∥ and friction f. Using Newton's 2d law, with subscripts to announce quantities parallel to the slope,

[latex]{a}_{\parallel }=\frac{{F}_{\text{net}\parallel }}{one thousand}[/latex]

where [latex]{F}_{\text{cyberspace}\parallel}={due west}_{\parallel}=mg\sin({25^{\circ}})[/latex], assuming no friction for this part, so that

[latex]a_{\parallel}=\frac{{F}_{\text{net}\parallel}}{k}=\frac{{mg}\sin({25}^{\circ})}{m}=yard\sin({25}^{\circ})[/latex]

(9.98 m/sⁱⁱ)(0.4226) = 4.14 m/s^two

is the acceleration.

(b) Including friction. Nosotros now have a given value for friction, and we know its management is parallel to the slope and it opposes motion between surfaces in contact. So the net external strength is now

[latex]{F}_{\text{net}\parallel }={westward}_{\parallel }-f[/latex]

and substituting this into Newton'southward 2nd police force, [latex]{a}_{\parallel}=\frac{{F}_{\text{net}\parallel }}{m}[/latex], gives

[latex]{a}_{\parallel}=\frac{{F}_{\text{net}\parallel}}{k}=\frac{{westward}_{\parallel}-f}{m}=\frac{{mg}\sin({25}^{\circ})-f}{m}[/latex].

We substitute known values to obtain

[latex]{a}_{\parallel }=\frac{(60.0\text{ kg})(ix.80\text{ chiliad/southward}^{2})(0.4226)-45.0\text{ North}}{60.0\text{ kg}}[/latex]

which yields

[latex]a_{\parallel}= iii.39\text{ thou/s}^{two}[/latex]

which is the acceleration parallel to the incline when there is 45.0 N of opposing friction.

Discussion

Since friction always opposes movement between surfaces, the dispatch is smaller when there is friction than when there is none. In fact, it is a full general result that if friction on an incline is negligible, then the acceleration down the incline isa =thou sinθ, regardless of mass. This is related to the previously discussed fact that all objects fall with the same acceleration in the absence of air resistance. Similarly, all objects, regardless of mass, slide down a frictionless incline with the same acceleration (if the angle is the aforementioned).

Resolving Weight into Components

Figure three. An object rests on an incline that makes an angle θ with the horizontal.

When an object rests on an incline that makes an bending θ with the horizontal, the force of gravity interim on the object is divided into two components: a force acting perpendicular to the plane, due west_⊥ , and a force acting parallel to the aeroplane,[latex]\textbf{westward}_{\parallel}[/latex]. The perpendicular force of weight, westward_⊥, is typically equal in magnitude and opposite in management to the normal forcefulness, N. The strength acting parallel to the plane, [latex]\textbf{w}_{\parallel}[/latex], causes the object to advance down the incline. The force of friction, f, opposes the motion of the object, and so it acts up forth the plane. Information technology is important to be conscientious when resolving the weight of the object into components. If the angle of the incline is at an angle θ to the horizontal, then the magnitudes of the weight components are

[latex]w_{\parallel}=w \sin{\theta} = mg \sin{\theta}[/latex]

and

[latex]w_{\perp}=w \cos{\theta} = mg \cos{\theta}[/latex]

Instead of memorizing these equations, it is helpful to be able to determine them from reason. To do this, describe the right triangle formed by the three weight vectors. Find that the angle θ of the incline is the same as the angle formed between w and w_⊥. Knowing this property, y'all tin can use trigonometry to decide the magnitude of the weight components:

[latex]\cos({\theta})=\frac{{westward}_{\perp}}{w}[/latex]

[latex]w_{\perp} = w \cos{\theta} = mg \cos({\theta})[/latex]

[latex]\sin{(\theta)}=\frac{{w}_{\parallel}}{westward}[/latex]

[latex]w_{\parallel} = w \sin{\theta} = mg \sin({\theta})[/latex]

Take-Habitation Experiment: Force Parallel

To investigate how a force parallel to an inclined plane changes, find a condom ring, some objects to hang from the end of the rubber band, and a board you can position at dissimilar angles. How much does the rubber ring stretch when y'all hang the object from the end of the lath? Now place the lath at an angle then that the object slides off when placed on the board. How much does the rubber band extend if it is lined up parallel to the board and used to hold the object stationary on the board? Try two more angles. What does this show?

Tension

A tension is a force along the length of a medium, especially a force carried by a flexible medium, such as a rope or cablevision. The word "tension" comes from a Latin word meaning "to stretch." Non coincidentally, the flexible cords that bear muscle forces to other parts of the trunk are called tendons. Any flexible connector, such as a string, rope, chain, wire, or cable, can exert pulls only parallel to its length; thus, a forcefulness carried by a flexible connector is a tension with management parallel to the connector. It is important to sympathize that tension is a pull in a connector. In contrast, consider the phrase: "You can't push a rope." The tension force pulls outward along the two ends of a rope. Consider a person belongings a mass on a rope as shown in Figure four.

Effigy iv. When a perfectly flexible connector (one requiring no force to curve information technology) such as this rope transmits a forcefulness T, that force must be parallel to the length of the rope, as shown. The pull such a flexible connector exerts is a tension. Note that the rope pulls with equal force just in opposite directions on the hand and the supported mass (neglecting the weight of the rope). This is an example of Newton'south third police force. The rope is the medium that carries the equal and contrary forces between the two objects. The tension anywhere in the rope between the paw and the mass is equal. Once yous have determined the tension in i location, you lot accept determined the tension at all locations along the rope.

Tension in the rope must equal the weight of the supported mass, equally we can prove using Newton'due south second police force. If the 5.00-kg mass in the effigy is stationary, then its acceleration is nil, and thus F_net= 0. The only external forces interim on the mass are its weight w and the tension T supplied by the rope. Thus,

F _net=T−westward= 0,

where T and due west are the magnitudes of the tension and weight and their signs bespeak direction, with up beingness positive here. Thus, only as you would wait, the tension equals the weight of the supported mass:

T = w = mg.

For a 5.00-kg mass, then (neglecting the mass of the rope) we run across that

T = mg = (5.00 kg)(9.80 m/south²) = 49.0 North

If we cutting the rope and insert a spring, the leap would extend a length corresponding to a force of 49.0 Northward, providing a direct observation and measure of the tension strength in the rope. Flexible connectors are ofttimes used to transmit forces around corners, such as in a hospital traction system, a finger articulation, or a bicycle restriction cable. If at that place is no friction, the tension is transmitted undiminished. Simply its management changes, and it is always parallel to the flexible connector. This is illustrated in Figure 5 (a) and (b).

Figure 5. (a) Tendons in the finger behave force T from the muscles to other parts of the finger, usually irresolute the force's direction, merely non its magnitude (the tendons are relatively friction costless). (b) The brake cable on a bicycle carries the tension T from the handlebars to the brake machinery. Again, the direction but non the magnitude of T is changed.

Case ii. What Is the Tension in a Tightrope?

Calculate the tension in the wire supporting the seventy.0-kg tightrope walker shown in Effigy vi.

Effigy half dozen. The weight of a tightrope walker causes a wire to sag past five.0 degrees. The arrangement of interest here is the point in the wire at which the tightrope walker is standing.

Strategy

Equally you lot can come across in the figure, the wire is not perfectly horizontal (information technology cannot be!), simply is aptitude under the person's weight. Thus, the tension on either side of the person has an upward component that tin back up his weight. As usual, forces are vectors represented pictorially by arrows having the same directions as the forces and lengths proportional to their magnitudes. The system is the tightrope walker, and the only external forces interim on him are his weight w and the two tensions T_Fifty (left tension) and T_R (right tension), as illustrated. It is reasonable to neglect the weight of the wire itself. The net external force is aught since the organisation is stationary. A little trigonometry can now exist used to find the tensions. I determination is possible at the outset—we tin see from part (b) of the figure that the magnitudes of the tensions T ₅₀ and T _R must be equal. This is because in that location is no horizontal acceleration in the rope, and the only forces acting to the left and right are T ₅₀ and T _R . Thus, the magnitude of those forces must be equal so that they cancel each other out.

Whenever we take 2-dimensional vector problems in which no two vectors are parallel, the easiest method of solution is to selection a convenient coordinate organisation and projection the vectors onto its axes. In this case the best coordinate system has one axis horizontal and the other vertical. Nosotros phone call the horizontal the x-centrality and the vertical the y-axis.

Solution

Offset, we demand to resolve the tension vectors into their horizontal and vertical components. It helps to describe a new free-trunk diagram showing all of the horizontal and vertical components of each force interim on the system.

Figure 7. When the vectors are projected onto vertical and horizontal axes, their components along those axes must add together to nothing, since the tightrope walker is stationary. The small bending results in T being much greater than w.

Consider the horizontal components of the forces (denoted with a subscript ten):

F _internetx = T _50x − T _Rx .

The net external horizontal strength F _{net x} = 0, since the person is stationary. Thus,

F _netx = T _Fiftyx − T _Rx

T _L10 =T _Rten

Now, observe Figure 7. Yous can use trigonometry to make up one's mind the magnitude of T _L and T _R. Detect that:

[latex]\brainstorm{array}{lll}{\cos}\left(5.0^{\circ}\right)& =& \frac{{T}_{\text{50}x}}{{T}_{\text{L}}}\\ {T}_{\text{Fifty}x}& =& {T}_{\text{50}}\cos\left(5.0^{\circ}\right)\\ \cos\left(five.0^{\circ}\right)& =& \frac{{T}_{\text{R}x}}{{T}_{\text{R}}}\\ {T}_{\text{R}10}& =& {T}_{\text{R}}\cos\left(5.0^{\circ}\right)\end{array}[/latex]

Equating T _{Fifty x} and T _Rx :

[latex]{T}_{\text{L}}\cos({5.0}^{\circ})={T}_{\text{R}}\cos({5.0}^{\circ})[/latex].

Thus,

T _Fifty=T _R=T,

as predicted. Now, because the vertical components (denoted past a subscript y), we tin can solve for T. Again, since the person is stationary, Newton'southward second constabulary implies that internet F _y =0. Thus, as illustrated in the free-body diagram in Effigy vii,

F _{net y} =T _Ly +T _Ry −w= 0.

Observing Figure 7, we can use trigonometry to make up one's mind the relationship betwixt T _Fiftyy , T _Ry , and T. As nosotros adamant from the analysis in the horizontal direction, T _L=T _R=T:

[latex]\begin{assortment}{lll}\sin\left(5.0^{\circ}\right)& =& \frac{{T}_{\text{L}y}}{{T}_{\text{L}}}\\ {T}_{\text{L}y}={T}_{\text{50}}\sin\left(5.0^{\circ} \right)& =& T\sin\left(v.0^{\circ}\right)\\ \sin\left(five.0^{\circ}\correct)& =& \frac{{T}_{\text{R}y}}{{T}_{\text{R}}}\\ {T}_{\text{R}y}={T}_{\text{R}}\sin\left(5.0^{\circ}\right)& =& T\sin\left(five.0^{\circ}\correct)\cease{assortment}[/latex].

Now, we can substitute the values for T _Fiftyy and T _Ry , into the net strength equation in the vertical management:

[latex]\begin{assortment}{lll}{F}_{\text{cyberspace}y}& =& {T}_{\text{L}y}+{T}_{\text{R}y}-w=0\\ {F}_{\text{cyberspace}y}& =& T\sin\left(5.0^{\circ}\correct)+T\sin\left(5.0^{\circ}\right)-w=0\\ 2T\sin\left(5.0^{\circ}\right)-west& =& 0\\ 2T\sin\left(five.0^{\circ} \right)& =& westward\end{assortment}[/latex]

and

[latex]T=\frac{w}{2\sin({5.0}^{\circ})}=\frac{\text{mg}}{ii\sin({v.0}^{\circ})}[/latex],

so that

[latex]T=\frac{(70.0\text{ kg})(9.80\text{ m/due south}^{2})}{ii(0.0872)}[/latex],

and the tension is

T= 3900 Northward.

Discussion

Annotation that the vertical tension in the wire acts as a normal strength that supports the weight of the tightrope walker. The tension is nigh half dozen times the 686-N weight of the tightrope walker. Since the wire is nearly horizontal, the vertical component of its tension is merely a minor fraction of the tension in the wire. The large horizontal components are in opposite directions and cancel, and and so virtually of the tension in the wire is not used to support the weight of the tightrope walker.

If we wish to create a very large tension, all nosotros have to practise is exert a forcefulness perpendicular to a flexible connector, every bit illustrated in Figure eight. Equally nosotros saw in the concluding case, the weight of the tightrope walker acted as a force perpendicular to the rope. We saw that the tension in the roped related to the weight of the tightrope walker in the following fashion:

[latex]T=\frac{w}{2\sin({\theta})}[/latex].

Nosotros can extend this expression to describe the tension T created when a perpendicular force (F_⊥) is exerted at the eye of a flexible connector:

[latex]T=\frac{{F}_{\perp }}{two\sin({\theta})}[/latex].

Note that θ is the angle betwixt the horizontal and the bent connector. In this case, T becomes very large as θ approaches nada. Even the relatively minor weight of any flexible connector will cause it to sag, since an infinite tension would result if information technology were horizontal (i.eastward., θ= 0 and sinθ = 0). (See Figure 8.)

Effigy 8. We tin can create a very large tension in the chain by pushing on it perpendicular to its length, as shown. Suppose we wish to pull a car out of the mud when no tow truck is available. Each fourth dimension the car moves forward, the chain is tightened to keep information technology equally about straight equally possible. The tension in the chain is given by [latex]T=\frac{{F}_{\perp }}{2\sin({\theta})}[/latex]; since θ is small, T is very large. This situation is analogous to the tightrope walker shown in Figure 6, except that the tensions shown hither are those transmitted to the motorcar and the tree rather than those acting at the point where F_⊥ is practical.

Figure nine. Unless an infinite tension is exerted, whatever flexible connector—such as the chain at the bottom of the movie—will sag under its own weight, giving a feature curve when the weight is evenly distributed along the length. Suspension bridges—such every bit the Aureate Gate Bridge shown in this image—are substantially very heavy flexible connectors. The weight of the span is evenly distributed along the length of flexible connectors, usually cables, which take on the characteristic shape. (credit: Leaflet, Wikimedia Commons)

Extended Topic: Real Forces and Inertial Frames

At that place is some other distinction among forces in improver to the types already mentioned. Some forces are real, whereas others are not. Real forces are those that accept some concrete origin, such as the gravitational pull. Contrastingly, fictitious forces are those that ascend simply considering an observer is in an accelerating frame of reference, such every bit ane that rotates (like a merry-go-round) or undergoes linear acceleration (like a car slowing down). For example, if a satellite is heading due due north in a higher place Earth'southward northern hemisphere, and then to an observer on Globe it will appear to experience a strength to the w that has no concrete origin. Of course, what is happening here is that Earth is rotating toward the east and moves east under the satellite. In Globe's frame this looks similar a westward force on the satellite, or it can be interpreted equally a violation of Newton's first law (the law of inertia). An inertial frame of reference is one in which all forces are real and, equivalently, i in which Newton's laws have the simple forms given in this chapter.

Earth'southward rotation is ho-hum plenty that Earth is most an inertial frame. You lot usually must perform precise experiments to notice fictitious forces and the slight departures from Newton'southward laws, such as the effect just described. On the large scale, such every bit for the rotation of weather systems and ocean currents, the effects can be easily observed.

The crucial factor in determining whether a frame of reference is inertial is whether it accelerates or rotates relative to a known inertial frame. Unless stated otherwise, all phenomena discussed in this text are considered in inertial frames.

All the forces discussed in this section are real forces, just at that place are a number of other existent forces, such as lift and thrust, that are non discussed in this section. They are more specialized, and information technology is not necessary to discuss every blazon of force. It is natural, however, to ask where the basic simplicity we seek to notice in physics is in the long listing of forces. Are some more bones than others? Are some unlike manifestations of the same underlying force? The answer to both questions is yes, as will exist seen in the side by side (extended) section and in the treatment of mod physics after in the text.

PhET Explorations: Forces in 1 Dimension

Explore the forces at work when you effort to push a filing cabinet. Create an applied forcefulness and run into the resulting friction strength and total force interim on the cabinet. Charts show the forces, position, velocity, and acceleration vs. fourth dimension. View a free-body diagram of all the forces (including gravitational and normal forces).

Click to download. Run using Java.

Section Summary

• When objects residual on a surface, the surface applies a strength to the object that supports the weight of the object. This supporting forcefulness acts perpendicular to and abroad from the surface. Information technology is chosen a normal force, N.
• When objects residual on a non-accelerating horizontal surface, the magnitude of the normal force is equal to the weight of the object:

  N = mg

• When objects rest on an inclined aeroplane that makes an angle θ with the horizontal surface, the weight of the object tin be resolved into components that deed perpendicular ([latex]{\mathbf{\text{w}}}_{\perp}[/latex]) and parallel ([latex]{\mathbf{\text{w}}}_{\parallel}[/latex]) to the surface of the plane. These components can exist calculated using:
  • [latex]{w}_{\parallel}=w\sin({\theta})=mg\sin({\theta})[/latex]

  • [latex]{w}_{\perp}=w\cos({\theta})=mg\cos({\theta})[/latex]

• The pulling force that acts along a stretched flexible connector, such every bit a rope or cable, is called tension, T. When a rope supports the weight of an object that is at remainder, the tension in the rope is equal to the weight of the object:

  T = mg.

• In any inertial frame of reference (i that is not accelerated or rotated), Newton's laws accept the simple forms given in this affiliate and all forces are real forces having a physical origin.

Conceptual Questions

one. If a leg is suspended past a traction setup as shown in Figure 9, what is the tension in the rope?

Figure 9. A leg is suspended by a traction system in which wires are used to transmit forces. Frictionless pulleys change the direction of the force T without changing its magnitude.

2. In a traction setup for a broken os, with pulleys and rope bachelor, how might nosotros be able to increase the force along the tibia using the same weight? (See Figure nine.) (Note that the tibia is the shin bone shown in this image.)

Bug & Exercises

1. 2 teams of nine members each engage in a tug of war. Each of the first team's members has an average mass of 68 kg and exerts an boilerplate force of 1350 North horizontally. Each of the second team's members has an boilerplate mass of 73 kg and exerts an average force of 1365 N horizontally. (a) What is magnitude of the acceleration of the two teams? (b) What is the tension in the section of rope betwixt the teams?

ii. What force does a trampoline have to apply to a 45.0-kg gymnast to accelerate her straight upwards at 7.50 m/due south²? Note that the reply is independent of the velocity of the gymnast—she tin can be moving either upward or down, or be stationary.

3. (a) Calculate the tension in a vertical strand of spider web if a spider of mass 8.00 × x^-5 hangs motionless on information technology. (b) Calculate the tension in a horizontal strand of spider web if the same spider sits motionless in the heart of information technology much similar the tightrope walker in Figure 6. The strand sags at an angle of 12º below the horizontal. Compare this with the tension in the vertical strand (discover their ratio).

iv. Suppose a 60.0-kg gymnast climbs a rope. (a) What is the tension in the rope if he climbs at a constant speed? (b) What is the tension in the rope if he accelerates upwards at a charge per unit of i.50 m/sⁱⁱ?

5. Testify that, as stated in the text, a force [latex]{\mathbf{\text{F}}}_{\perp}[/latex] exerted on a flexible medium at its heart and perpendicular to its length (such as on the tightrope wire in Figure six) gives rise to a tension of magnitude [latex]T=\frac{{F}_{\perp }}{two\sin(\theta)}[/latex].

six. Consider the baby beingness weighed in Figure ten. (a) What is the mass of the child and basket if a scale reading of 55 N is observed? (b) What is the tension T ₁ in the cord attaching the babe to the scale? (c) What is the tension T ₂ in the string attaching the scale to the ceiling, if the scale has a mass of 0.500 kg? (d) Draw a sketch of the state of affairs indicating the system of interest used to solve each part. The masses of the cords are negligible.

Figure 10. A baby is weighed using a bound calibration.

Glossary

inertial frame of reference:

a coordinate system that is not accelerating; all forces interim in an inertial frame of reference are real forces, as opposed to fictitious forces that are observed due to an accelerating frame of reference

normal force:

the force that a surface applies to an object to support the weight of the object; acts perpendicular to the surface on which the object rests

tension:

the pulling forcefulness that acts forth a medium, especially a stretched flexible connector, such every bit a rope or cable; when a rope supports the weight of an object, the force on the object due to the rope is called a tension strength

Selected Solutions to Issues & Exercises

1. (a) 0.eleven m/s² (b) ane.two × 10^four North

iii. (a) 7.84 × 10^-4 Northward (b) 1.89 × 10^-3 Northward. This is 2.41 times the tension in the vertical strand.

five.  Newton's second law applied in vertical direction gives

[latex]{F}_{y}=F - 2 T\sin{\theta}=0[/latex]

[latex]{F}=2 T\sin{\theta}[/latex]

[latex]T=\frac{F}{2\sin{\theta}}[/latex].

Source: https://courses.lumenlearning.com/physics/chapter/4-5-normal-tension-and-other-examples-of-forces/

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