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Chapter 3: Problem 11

A \(5.0\) -kg object is to be given an upward acceleration of \(0.30 \mathrm{~m} / \mathrm{s}^{2}\) by a rope pulling straight upward on it. What must be the tension in the rope? The free-body diagram for the object is shown in Fig. \(3-8(b)\). The tension in the rope is \(F_{T}\), and the weight of the object is \(F_{W}=m g=(5.0 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)=49.1 \mathrm{~N}\). Using \(\sum F_{y}=m a_{y}\) with \(u p\) taken as positive, $$ +\uparrow \sum F_{y}=F_{T}-m g=m a_{y} \quad \text { or } \quad F_{T}-49.1 \mathrm{~N}=(5.0 \mathrm{~kg})\left(0.30 \mathrm{~m} / \mathrm{s}^{2}\right) $$ from which \(F_{T}=50.6 \mathrm{~N}=51 \mathrm{~N}\). As a check, we notice that \(F_{T}\) is larger than \(F_{W}\), as it must be if the object is to accelerate upward.

Short Answer

Expert verified
The tension in the rope must be approximately 51 N.

Step by step solution

01

Identify Known Variables

We know the following:- Mass of the object, \( m = 5.0 \) kg- Acceleration due to gravity, \( g = 9.81 \text{ m/s}^2 \)- Desired upward acceleration, \( a = 0.30 \text{ m/s}^2 \)
02

Calculate the Weight of the Object

The weight \( F_W \) of the object is given by \( F_W = m \cdot g \).\[F_W = 5.0 \text{ kg} \cdot 9.81 \text{ m/s}^2 = 49.1 \text{ N}\]
03

Apply Newton's Second Law

According to Newton's second law, the sum of the forces in the \( y \)-direction is \( \Sigma F_y = m \cdot a \).We are given:\[\Sigma F_y = F_T - F_W\]Where \( F_T \) is the tension in the rope.
04

Substitute and Solve for Tension

Replace \( F_W \) and solve for \( F_T \):\[F_T - 49.1 \text{ N} = 5.0 \text{ kg} \cdot 0.30 \text{ m/s}^2\]\[F_T - 49.1 \text{ N} = 1.5 \text{ N}\]Adding \( 49.1 \text{ N} \) to both sides gives:\[F_T = 50.6 \text{ N}\]
05

Check the Result

Since \( F_T = 50.6 \text{ N} \) is greater than the weight \( F_W = 49.1 \text{ N} \), it implies that the object will have the desired upward acceleration, validating our calculation.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Tension in Physics
Tension is the force exerted along a rope, string, or similar object when it is pulled tight by forces acting on opposite ends. In our exercise, tension is what the rope needs to exert to lift the object upwards with the desired acceleration. When a force is applied to a rope to lift an object, the tension force will oppose the weight of the object due to gravity.
To find the tension in a rope lifting a 5.0 kg object, we need to consider both the gravitational force acting downwards and the force needed to accelerate the object upwards. This means the tension in the rope must be greater than the weight of the object to achieve the required upward acceleration. By calculating this tension, we can determine how strong the rope's pulling force must be to fulfill the problem鈥檚 conditions.
The Role of a Free-Body Diagram
A free-body diagram is an essential tool in physics used to visualize the forces acting on an object. It breaks down complex interactions into simpler components and helps in understanding how different forces like tension, gravity, and applied forces interact with one another.
In our example, the free-body diagram illustrates the tension force exerted by the rope (denoted as \(F_T\)), and the downward force due to the object鈥檚 weight (\(F_W = mg\)). By setting up a diagram, it becomes clearer how these forces play roles in determining the net force and resulting motion. The free-body diagram helps us apply Newton鈥檚 Second Law effectively by visually showing the direction and magnitude of each force involved.
Calculating Acceleration Using Newton's Second Law
Newton's Second Law states that the acceleration of an object depends on the net force acting on it and on its mass \((\sum F = ma)\). In other words, how fast an object moves depends on how much force is applied to it and how heavy it is.
For the given problem, we are provided with certain forces: tension in the rope and the gravitational force pulling the object down. The net force acting upward can be considered as the difference between the upward tension force \(F_T\) and the gravitational force \(F_W\). The formula for calculating tension based on desired acceleration is given by:
* \(F_T - F_W = m \cdot a\)
* Where \(F_W\) is the object's weight and \(a\) is the required upward acceleration
By substituting the known values of weight and desired acceleration, we solve for \(F_T\), which results in the necessary tension force needed to achieve the desired motion. This shows how the right application of Newton鈥檚 Second Law allows us to solve real-world physics problems with precise calculations.

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Most popular questions from this chapter

A horizontal force of \(140 \mathrm{~N}\) is needed to pull a \(60.0-\mathrm{kg}\) box across a horizontal floor at constant speed. What is the coefficient of friction between floor and box? Determine it to three significant figures even though that's quite unrealistic. The free-body diagram for the box is rendered in Fig. \(3-9 .\) Because the box does not move up or down, \(a_{y}=0\). Therefore, $$ +\uparrow \sum F_{y}=m a_{y} \quad \text { yields } \quad F_{N}-m g=(m)\left(0 \mathrm{~m} / \mathrm{s}^{2}\right) $$ from which we find that \(F_{N}=m g=(60.0 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)=588.6 \mathrm{~N}\). Further, because the box is moving horizontally at constant speed, \(a_{x}=0\) and so $$ \pm \sum F_{x}=m a_{x} \quad \text { leads to } \quad 140 \mathrm{~N}-F_{\mathrm{f}}=0 $$ from which the friction force is \(F_{\mathrm{f}}=140 \mathrm{~N}\). Then $$ \mu_{k}=\frac{F_{\mathrm{f}}}{F_{N}}=\frac{140 \mathrm{~N}}{588.6 \mathrm{~N}}=0.238 $$

A \(20.0 \mathrm{~kg}\) object that can move freely is subjected to a resultant force of \(45.0 \mathrm{~N}\) in the \(-x\) -direction. Find the acceleration of the object. We make use of the second law in component form, \(\sum F_{x}=m a_{x}\), with \(\sum F_{x}=-45.0 \mathrm{~N}\) and \(m=20.0 \mathrm{~kg}\). Then $$ a_{x}=\frac{\sum F_{x}}{m}=\frac{-45.0 \mathrm{~N}}{20.0 \mathrm{~kg}}=-2.25 \mathrm{~N} / \mathrm{kg}=-2.25 \mathrm{~m} / \mathrm{s}^{2} $$ where we have used the fact that \(1 \mathrm{~N}=1 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}^{2}\). Because the resultant force on the object is in the \(-x\) -direction, its acceleration is also in that direction.

A car whose weight is \(F_{W}\) is on a ramp which makes an angle \(\theta\) to the horizontal. How large a perpendicular force must the ramp withstand if it is not to break under the car's weight? As rendered in Fig. \(3-6\), the car's weight is a force \(\overrightarrow{\mathbf{F}}_{W}\) that pulls straight down on the car. We take components of \(\overrightarrow{\mathbf{F}}\) along the incline and perpendicular to it. The ramp must balance the force component \(F_{W} \cos \theta\) if the car is not to crash through the ramp. In other words, the force exerted on the car by the ramp, upwardly perpendicular to the ramp, is \(F_{N}\) and \(F_{N}=F_{W} \cos \theta\).

A 20 -kg crate hangs at the end of a long rope. Find its acceleration (magnitude and direction) when the tension in the rope is \((a) 250 \mathrm{~N},(b) 150 \mathrm{~N},(c)\) zero, \((d) 196 \mathrm{~N}\).

A child pulls on a rope attached to a sled with a force of \(60 \mathrm{~N}\). The rope makes an angle of \(40^{\circ}\) to the ground. ( \(a\) ) Compute the effective value of the pull tending to move the sled along the ground. (b) Compute the force tending to lift the sled vertically. As depicted in Fig. \(3-5\), the components of the \(60 \mathrm{~N}\) force are \(39 \mathrm{~N}\) and \(46 \mathrm{~N}\). ( \(a\) ) The pull along the ground is the horizontal component, \(46 \mathrm{~N}\). (b) The lifting force is the vertical component, \(39 \mathrm{~N}\).
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