/*! 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 41 A large crate with mass \(m\) re... [FREE SOLUTION] | 91Ó°ÊÓ

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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 \(\mu_{\mathrm{s}}\)

Short Answer

Expert verified
The magnitude of the force required to keep the crate moving at constant velocity can be solved from the equation \(F = \frac{\mu_{k}(mg - Fsin\theta)}{cos\theta}\). The critical value for static friction \(\mu_{s}\) is given by \(\mu_{s} > tan\theta\).

Step by step solution

01

STEP 1: IDENTIFY THE FORCES ACTING ON THE CRATE

There are four forces acting on the crate: The gravitational force (downward), the normal force (upward), frictional force (horizontal and opposing motion), and applied force by the woman (at an angle \(\theta\) below the horizontal). The gravitational force is \(mg\), the normal force is \(N\), the frictional force is \(\mu_{k}N\) for part (a) and \(\mu_{s}N\) for part (b), and the applied force is \(F\).
02

STEP 2: RESOLVE THE FORCES INTO COMPONENTS

Since the forces act in different directions, resolve the forces into vertical and horizontal components. The vertical component of the applied force is \(Fsin\theta\) (downward) and the horizontal component is \(Fcos\theta\) (to the right).
03

STEP 3: SET UP THE EQUATIONS

To keep the crate moving at a constant velocity, the net force acting on the crate must be zero. Hence, for vertical forces: \(N + Fsin\theta = mg\), and for horizontal forces: \(Fcos\theta = \mu_{k}N\).
04

STEP 4: SOLVE FOR THE REQUIRED FORCE

Solve the set of equations from Step 3. First, find the normal force \(N= mg - Fsin\theta\) from the vertical forces equation. Substitute \(N\) in the horizontal equation to find the force \(F = \frac{\mu_{k}(mg - Fsin\theta)}{cos\theta}\). Solve this equation for \(F\) to determine the magnitude of force required to keep the crate moving at a constant velocity.
05

STEP 5: DETERMINE THE CRITICAL VALUE OF STATIC FRICTION

In order for the crate to start moving, the horizontal component of the applied force must be greater than the static friction. Hence, the inequality is \(Fcos\theta > \mu_{s}N\). Using \(N = mg- Fsin\theta\) and after some algebraic manipulations, the critical value of static friction \(\mu_{s}\) is found to be: \(\mu_{s} > tan\theta\).

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Key Concepts

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

Frictional Force
Understanding the frictional force is essential when it comes to analyzing the movement of objects. It is the force exerted by a surface as an object moves across it or makes an effort to move across it. In the context of the crate on the floor, there are two types of friction to consider: static friction and kinetic friction.

Static friction is at play when the crate is not moving, and it must be overcome for the crate to start moving. Once the crate is in motion, kinetic friction, also known as sliding or dynamic friction, takes over. This force is usually less than the maximum static friction, making it easier to keep an object moving than to start it moving. The force of kinetic friction (\f\(\f\rmu_{k}N\f\)) is calculated by multiplying the coefficient of kinetic friction (\f\(\f\rmu_{k}\f\)) by the normal force (\f\(N\f\)) exerting on the crate from the surface it's on.
Net Force
Net force is the sum of all the forces acting on an object in any given situation. According to Newton's second law of motion, the net force is equal to the mass of the object multiplied by its acceleration (\f\(F_{net} = m \times a\f\)). If the crate is moving at a constant velocity, this implies that the crate's acceleration is zero. Consequently, the net force is also zero. This translates to the forces being balanced, with the force exerted by the woman pushing the crate being exactly counterbalanced by the force of friction. While calculating net force, it's crucial to consider both the magnitude and direction of all forces involved and sum them vectorially.
Constant Velocity Motion
Constant velocity motion occurs when an object moves with a constant speed and in a straight line. For an object to maintain constant velocity, the net force acting on it must be zero. This state is a key principle of Newton's first law of inertia, which states that an object will stay at rest or in uniform motion unless acted upon by an external force. In the exercise's scenario, for the crate to move with constant velocity, the woman needs to apply a force that perfectly counters the kinetic friction. Hence, we need to ensure that the applied horizontal force (\f\(Fcos(\theta)\f\)) matches the force of friction (\f\(\f\rmu_{k}N\f\)) to achieve constant velocity motion.
Critical Value of Static Friction
The critical value of static friction is the threshold at which the static friction force transitions from preventing an object from moving to allowing it to move. It is dependent on both the surfaces in contact and the normal force, and it represents the maximum force that needs to be overcome to initiate movement. Typically, the critical value is expressed as the coefficient of static friction (\f\(\f\rmu_{s}\f\)), which when multiplied by the normal force, gives the maximum static frictional force. As stated in the solution, the woman can't move the crate if the static friction coefficient exceeds a certain limit, specifically if \f\(\f\rmu_{s} > tan(\theta)\f\). This relationship shows the importance of both the angle of application and the surface characteristics when determining the effort required to start moving an object.

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Most popular questions from this chapter

A box with mass \(10.0 \mathrm{~kg}\) moves on a ramp that is inclined at an angle of \(55.0^{\circ}\) above the horizontal. The coefficient of kinetic friction between the box and the ramp surface is \(\mu_{\mathrm{k}}=0.300 .\) Calculate the magnitude of the acceleration of the box if you push on the box with a constant force \(F=120.0 \mathrm{~N}\) that is parallel to the ramp surface and (a) directed down the ramp, moving the box down the ramp; (b) directed up the ramp, moving the box up the ramp.

Force on a Skater's Wrist. A \(52 \mathrm{~kg}\) ice skater spins about a vertical axis through her body with her arms horizontally outstretched; she makes 2.0 turns each second. The distance from one hand to the other is \(1.50 \mathrm{~m} .\) Biometric measurements indicate that each hand typically makes up about \(1.25 \%\) of body weight. (a) Draw a free-body diagram of one of the skater's hands. (b) What horizontal force must her wrist exert on her hand? (c) Express the force in part (b) as a multiple of the weight of her hand.

A bowling ball weighing \(71.2 \mathrm{~N}(16.0 \mathrm{lb})\) is attached to the ceiling by a \(3.80 \mathrm{~m}\) rope. The ball is pulled to one side and released; it then swings back and forth as a pendulum. As the rope swings through the vertical, the speed of the bowling ball is \(4.20 \mathrm{~m} / \mathrm{s}\). At this instant, what are (a) the acceleration of the bowling ball, in magnitude and direction, and (b) the tension in the rope?

Two ropes are connected to a steel cable that supports a hanging weight (Fig. P5.59). (a) Draw a freebody diagram showing all of the forces acting at the knot that connects the two ropes to the steel cable. Based on your diagram, which of the two ropes will have the greater tension? (b) If the maximum tension either rope can sustain without breaking is \(5000 \mathrm{~N},\) determine the maximum value of the hanging weight that these ropes can safely support. Ignore the weight of the ropes and of the steel cable.

If the coefficient of static friction between a table and a uniform, massive rope is \(\mu_{\mathrm{s}},\) what fraction of the rope can hang over the edge of the table without the rope sliding?

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