/*! 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 47 \(\bullet$$\bullet\) A \(25.0-\m... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

\(\bullet$$\bullet\) A \(25.0-\mathrm{kg}\) box of textbooks rests on a loading ramp that makes an angle \(\alpha\) with the horizontal. The coefficient of kinetic friction is \(0.25,\) and the coefficient of static friction is 0.35 . (a) As the angle \(\alpha\) is increased, find the minimum angle at which the box starts to slip. (b) At this angle, find the accel- eration once the box has begun to move. (c) At this angle, how fast will the box be moving after it has slid 5.0 \(\mathrm{m}\) along the loading ramp"?

Short Answer

Expert verified
(a) The box starts to slip at \( \alpha \approx 19.29^\circ \). (b) Acceleration is \( 1.33 \, \text{m/s}^2 \). (c) Final speed after sliding 5 m is \( 3.65 \, \text{m/s} \).

Step by step solution

01

Analyze the forces at play

When the box just starts to slip, the static friction force is at its maximum and equals the component of gravity along the ramp.The force of gravity acting down the ramp is given by \[ F_{g, parallel} = m imes g imes \sin(\alpha) \]The maximum static friction force is \[ f_{s, max} = \mu_s \times N \]where the normal force \( N \) is \[ N = m \times g \times \cos(\alpha) \] for the weight component perpendicular to the ramp.
02

Determine the angle at which the box starts to slip

Set the forces equal since the static friction force will equal the gravitational force component when slipping starts:\[ \mu_s \times m \times g \times \cos(\alpha) = m \times g \times \sin(\alpha) \]Cancel out the common terms and solve for \( \alpha \):\[ \mu_s \times \cos(\alpha) = \sin(\alpha) \]Divide by \( \cos(\alpha) \):\[ \tan(\alpha) = \mu_s \]Substitute \( \mu_s = 0.35 \):\[ \alpha = \tan^{-1}(0.35) \approx 19.29^\circ \]
03

Calculate the acceleration of the box

Once the box starts to move, the kinetic friction takes over. The force of gravity down the ramp minus kinetic friction gives the net force:\[ F_{net} = m \times g \times \sin(\alpha) - \mu_k \times m \times g \times \cos(\alpha) \]Using \( F = ma \):\[ a = g \times (\sin(\alpha) - \mu_k \times \cos(\alpha)) \]Substitute the known values:\[ a = 9.81 \times (\sin(19.29^\circ) - 0.25 \times \cos(19.29^\circ)) \]Calculate:\[ a \approx 1.33 \, \text{m/s}^2 \]
04

Determine final speed after sliding down the ramp

Use the kinematic equation for velocity when starting from rest:\[ v^2 = u^2 + 2as \]Here, initial speed \( u = 0 \), acceleration \( a \approx 1.33 \, \text{m/s}^2 \), and distance \( s = 5.0 \, \text{m} \):\[ v^2 = 2 \times 1.33 \times 5.0 \]Calculate for \( v \):\[ v = \sqrt{2 \times 1.33 \times 5.0} \approx 3.65 \, \text{m/s} \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Static Friction
Static friction is the force that keeps an object at rest when it is subject to external forces. It acts in the opposite direction to the initiating force. For an object like a box on a ramp, static friction prevents it from sliding down until the ramp is tilted sufficiently. The maximum static friction is determined by the formula:\[ f_{s, max} = \mu_s \times N \]Here, \( \mu_s \) is the coefficient of static friction and \( N \) is the normal force, which is the perpendicular force exerted by a surface on the object. In the scenario with the textbook box, as the ramp angle increases, the component of gravitational force parallel to the ramp also increases. When this component equals the maximum static frictional force, the box starts to slip.
Kinetic Friction
Once the box starts moving, static friction is no longer acting. Instead, kinetic friction comes into play. This is usually less than static friction, which means it's easier to keep an object moving than to start moving it. The kinetic friction force is determined by:\[ f_k = \mu_k \times N \]where \( \mu_k \) is the coefficient of kinetic friction. For our box, after it starts slipping, kinetic friction acts against the motion as it slides down the ramp. This opposing force is crucial in calculating the net force and resulting acceleration when the box is in motion. Knowing the kinetic friction also helps us predict how quickly the box will reach the bottom of the ramp.
Force Analysis
In this context, force analysis involves breaking down the forces acting on the box into components parallel and perpendicular to the ramp. The gravitational force can be split into:
  • Parallel component: \( F_{g, parallel} = m \times g \times \sin(\alpha) \)
  • Perpendicular component: \( F_{g, perpendicular} = m \times g \times \cos(\alpha) \)
Where \( m \) is the mass of the box, \( g \) is the acceleration due to gravity, and \( \alpha \) is the angle of the ramp. This analysis helps us understand how different forces balance each other out, and how to determine the angle at which static friction gives way to motion. Once the box moves, we consider net force and acceleration using these components and the friction forces.
Angle of Inclination
The angle of inclination \( \alpha \) is key in determining when the box will start to slip. It is the angle between the ramp and the horizontal ground. The calculation for the critical angle uses the relation:\[ \tan(\alpha) = \mu_s \]For our scenario, substituting \( \mu_s = 0.35 \), we calculate \( \alpha = \tan^{-1}(0.35) \approx 19.29^\circ \). This angle signifies the point where the component of gravitational force just matches the maximum static friction, leading the box to start slipping. Understanding this angle is crucial for predicting and controlling motion in physics problems involving inclined surfaces. Once the box begins to move, the dynamics shift to dealing with kinetic friction and acceleration.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

\(\bullet$$\bullet\) Prevention of hip injuries. People (especially the elderly) who are prone to falling can wear hip pads to cushion the impact on their hip from a fall. Experiments have shown that if the speed at impact can be reduced to 1.3 \(\mathrm{m} / \mathrm{s}\) or less, the hip will usually not fracture. Let us investigate the worst-case sce- nario, in which a 55 kg person completely loses her footing (such as on icy pavement) and falls a distance of \(1.0 \mathrm{m},\) the dis- tance from her hip to the ground. We shall assume that the per- son's entire body has the same acceleration, which, in reality, would not quite be true. (a) With what speed does her hip reach the ground? (b) A typical hip pad can reduce the person's speed to 1.3 \(\mathrm{m} / \mathrm{s}\) over a distance of 2.0 \(\mathrm{cm} .\) Find the acceleration (assumed to be constant) of this person's hip while she is slow- ing down and the force the pad exerts on it. (c) The force in part (b) is very large. To see if it is likely to cause injury, calcu- late how long it lasts.

\(\bullet$$\bullet\) The monkey and her bananas. \(\Lambda 20 \mathrm{kg}\) monkey has a firm hold on a light rope that passes over a frictionless pulley and is attached to a 20 \(\mathrm{kg}\) bunch of bananas (Figure 5.72 ). The monkey looks up, sees the bananas, and starts to climb the rope to get them. (a) As the monkey climbs, do the bananas move up, move down, or remain at rest? (b) As the monkey climbs, does the distance between the monkey and the ba- nanas decrease, increase, or remain con- stant? (c) The monkey releases her hold on - the rope. What happens to the distance be- tween the monkey and the bananas while she is falling? (d) Before reaching the ground, the monkey grabs the rope to stop her fall. What do the bananas do?

\(\bullet$$\bullet\) Stopping distance of a car. (a) If the coefficient of kinetic friction between tires and dry pavement is \(0.80,\) what is the shortest distance in which you can stop an automobile by lock- ing the brakes when traveling at 29.1 \(\mathrm{m} / \mathrm{s}\) (about 65 \(\mathrm{mi} / \mathrm{h} )\) ? (b) On wet pavement, the coefficient of kinetic friction may be only \(0.25 .\) How fast should you drive on wet pavement in ordel to be able to stop in the same distance as in part (a)? (Note. Locking the brakes is not the safest way to stop.)

\(\bullet$$\bullet\) A 65.0 -kg parachutist falling vertically at a speed of 6.30 \(\mathrm{m} / \mathrm{s}\) impacts the ground, which brings him to a complete stop in a distance of 0.92 \(\mathrm{m}\) (roughly half of his height). Assuming constant acceleration after his feet first touch the ground, what is the average force exerted on the parachutist by the ground?

\(\bullet$$\bullet\) A toboggan approaches a snowy hill moving at 11.0 \(\mathrm{m} / \mathrm{s}\) . The coefficients of static and kinetic friction between the snow and the tobogan are 0.40 and 0.30 , respectively, and the hill slopes upward at \(40.0^{\circ}\) above the horizontal. Find the acceleration of the tobogan (a) as it is going up the hill and (b) after it has reached its highest point and is sliding down the hill.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.