/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 38 A box with mass \(m\) is dragged... [FREE SOLUTION] | 91Ó°ÊÓ

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A box with mass \(m\) is dragged across a level floor with coefficient of kinetic friction \(\mu_{\mathrm{k}}\) by a rope that is pulled upward at an angle \(\theta\) above the horizontal with a force of magnitude \(F\). (a) In terms of \(m, \mu_{\mathrm{k}}, \theta,\) and \(g,\) obtain an expression for the magnitude of the force required to move the box with constant speed. (b) Knowing that you are studying physics, a CPR instructor asks you how much force it would take to slide a \(90 \mathrm{~kg}\) patient across a floor at constant speed by pulling on him at an angle of \(25^{\circ}\) above the horizontal. By dragging weights wrapped in an old pair of pants down the hall with a spring balance, you find that \(\mu_{\mathrm{k}}=0.35 .\) Use the result of part (a) to answer the instructor's question.

Short Answer

Expert verified
A force of approximately 385 N is required.

Step by step solution

01

Analysis of forces

Identify all the forces on the box: There is a downwards force due to gravity equal to \(mg\). There is a normal force upwards from the ground. The kinetic friction force \(F_{k}\) opposes the motion of the box and there is an applied force \(F\) at an angle \(\theta\) above the horizontal.
02

Newton’s second law in the vertical direction

According to Newton’s second law, the sum of the forces in the vertical direction is zero, because the box is not moving vertically. So, we write \(N - mg + F \sin \theta = 0\). From this equation, we find the normal force \(N = mg - F\sin\theta\)
03

Newton’s second law in the horizontal direction

Again, the sum of the forces in the horizontal direction is zero, because the box is moving with a constant speed (no acceleration). The friction force is given by \(F_{k} = \mu_k N\). So we write \(- F_{k} + F \cos \theta = 0\), and replace \(F_{k}\) and \(N\) with their expressions from above. This gives \(F = \frac{ mg \mu_k}{\cos\theta + \mu_k \sin\theta}\)
04

Substitute numbers from part (b)

To estimate the necessary force to move the patient, plug in the given values \(m = 90\, \mathrm{kg}\), \(\mu_{\mathrm{k}} = 0.35\), \(\theta = 25^{\circ}, g = 9.8\, \mathrm{m/s^2}\) in the formula from step 3. Note that you need to convert the angle from degree to radian before substituting it.
05

Calculation

After substituting the given values, the formula gives \(F \approx 385 \mathrm{N}\). Therefore, it would take a force of approximately 385 N to drag the patient across the floor at a constant speed.

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Key Concepts

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

Newton's Second Law
Newton's Second Law plays a crucial role in understanding how forces affect motion. It states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. This can be expressed through the equation: \[ F_{ ext{net}} = ma \] where \( F_{\text{net}} \) is the net force, \( m \) is the mass, and \( a \) is the acceleration of the object. In this exercise, however, we focused on the scenario where the net force, particularly in the vertical and horizontal directions, is zero. This is because the box is moving at a constant speed and not accelerating. For the vertical direction, we applied Newton's Second Law by setting the sum of forces to zero: \[ N - mg + F \sin \theta = 0 \] This ensures that the vertical forces are balanced since the box doesn’t move up or down. Similarly, in the horizontal direction, we balance forces to achieve constant speed by:\[ - F_{k} + F \cos \theta = 0 \] This states that the horizontal applied force is perfectly countered by the opposing friction force.
Normal Force
The normal force, denoted as \( N \), is the force exerted by a surface to support the weight of an object resting on it, acting perpendicular to the surface. In the context of this problem, the normal force counteracts gravity and the component of the applied force, which tries to lift the box off the ground.The normal force is crucial because it affects the frictional force, which relies on the normal force's magnitude. From the vertical forces balance, as per Newton's Second Law:\[ N = mg - F \sin \theta \] Here, \( mg \) is the gravitational force, and \( F \sin \theta \) is the upward force component. This formula shows that as the angle \( \theta \) increases, resulting in a larger upward component, the normal force decreases, reducing the frictional force.Normal force plays an integral role when calculating the kinetic friction, as friction depends on this adaptive force.
Constant Speed Motion
Constant speed motion implies there is no acceleration. In practical terms, this means the forces acting on the object must be perfectly balanced out.Since the object's speed remains unchanged, according to Newton's Second Law, the net horizontal force is zero. This is because:\[ F_{\text{applied}} = F_{k} \]Where \( F_{\text{applied}} \) is the part of the force that aims to pull the box horizontally, represented by \( F \cos \theta \). The friction force \( F_{k} \) opposes this motion. Hence:\[ F \cos \theta = mg \mu_k \]When moving at constant speed, forces in both the horizontal and vertical directions must sum to zero. Thus, the applied force must exactly match the kinetic friction force to maintain that speed. This principle helped derive the formula necessary to exert just the right force to move the box without accelerating.

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Most popular questions from this chapter

A block with mass \(m_{1}\) is placed on an inclined plane with slope angle \(\alpha\) and is connected to a hanging block with mass \(m_{2}\) by a cord passing over a small, friction less pulley (Fig. P5.74). The coefficient of static friction is \(\mu_{\mathrm{s}}\), and the coefficient of kinetic friction is \(\mu_{\mathrm{k}}\). (a) Find the value of \(m_{2}\) for which the block of mass \(m_{1}\) moves up the plane at constant speed once it is set in motion. (b) Find the value of \(m_{2}\) for which the block of mass \(m_{1}\) moves down the plane at constant speed once it is set in motion. (c) For what range of values of \(m_{2}\) will the blocks remain at rest if they are released from rest?

A large crate with mass \(m\) rests on a horizontal floor. The coefficients of friction between the crate and the floor are \(\mu_{\mathrm{s}}\) and \(\mu_{\mathrm{k}} .\) A woman pushes downward with a force \(\overrightarrow{\boldsymbol{F}}\) on the crate at an angle \(\theta\) below the horizontal. (a) What magnitude of force \(\vec{F}\) is required to keep the crate moving at constant velocity? (b) If \(\mu_{\mathrm{s}}\) is greater than some critical value, the woman cannot start the crate moving no matter how hard she pushes. Calculate this critical value of \(\mu_{\mathrm{s}}\)

Jack sits in the chair of a Ferris wheel that is rotating at a constant \(0.100 \mathrm{rev} / \mathrm{s}\). As Jack passes through the highest point of his circular path, the upward force that the chair exerts on him is equal to one- fourth of his weight. What is the radius of the circle in which Jack travels? Treat him as a point mass.

DATA In your physics lab, a block of mass m is at rest on a horizontal surface. You attach a light cord to the block and apply a horizontal force to the free end of the cord. You find that the block remains at rest until the tension \(T\) in the cord exceeds \(20.0 \mathrm{~N}\). For \(T>20.0 \mathrm{~N},\) you measure the acceleration of the block when \(T\) is maintained at a constant value, and you plot the results (Fig. \(\mathrm{P} 5.109)\). The equation for the straight line that best fits your data is \(a=\left[0.182 \mathrm{~m} /\left(\mathrm{N} \cdot \mathrm{s}^{2}\right)\right] T-2.842 \mathrm{~m} / \mathrm{s}^{2}\) For this block and surface, what are (a) the coefficient of static friction and (b) the coefficient of kinetic friction? (c) If the experiment were done on the earth's moon, where \(g\) is much smaller than on the earth, would the graph of \(a\) versus \(T\) still be fit well by a straight line? If so, how would the slope and intercept of the line differ from the values in Fig. \(\mathrm{P} 5.109 ?\) Or, would each of them be the same?

An \(8.00 \mathrm{~kg}\) block of ice, released from rest at the top of a 1.50-m-long friction less ramp, slides downhill, reaching a speed of \(2.50 \mathrm{~m} / \mathrm{s}\) at the bottom. (a) What is the angle between the ramp and the horizontal? (b) What would be the speed of the ice at the bottom if the motion were opposed by a constant friction force of \(10.0 \mathrm{~N}\) parallel to the surface of the ramp?

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