/*! 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} Q13Q A box is on a ramp that is at an... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A box is on a ramp that is at angleθto the horizontal. Asθ is increased from zero, and before the box slips, do the following increase, decrease, or remain the same: (a) the component of the gravitational force on the box, along the ramp, (b) the magnitude of the static frictional force on the box from the ramp, (c) the component of the gravitational force on the box, perpendicular to the ramp, (d) the magnitude of the normal force on the box from the ramp, and (e) the maximum valuefs,max of the static frictional force?

Short Answer

Expert verified

a) The component of the gravitational force on the box, along with the ramp increases.

b) The magnitude of the static frictional force on the box from the ramp decreases.

c) The component of the gravitational force on the box, perpendicular to the ramp decreases.

d) The magnitude of the normal force on the box from the ramp

e) The maximum value of the static frictional force remains the same.

Step by step solution

01

The given data

(a) A box is on a ramp at an angle to the horizontal.

(b) The angle is increased from zero.

02

Understanding the concept of the force and friction

To observe which force will increase, decrease or remain the same while increasing the angle from 0, we have to draw the free body diagram of the box on the ramp and then apply Newton's 2nd law of motion. This determines the behavior upon the application of net forces acting on the body.

Formulae:

The force according to Newton’s second law,

F=ma (1)

The gravitational force acting on a body,

Fg=mg (2)

The static frictional force acting on a body,

fs=μ±· (3)

03

a) Calculation of the component of the gravitational force along the ramp

Free body diagram of the box on-ramp:


The component of the gravitational force along the ramp using equation (2) is given by, the Mgsinθ function increases with an increase in the angle θ.

Hence, the gravitational force component along the ramp increases with an increase in the angle.

04

b) Calculation of the magnitude of the static frictional force

To find the static frictional force on the box we have to use Newton’s 2nd law of motion along the vertical direction, we use equation (i) for the net forces as follows:

(Here, there is no motion along the vertical direction so a=0m/s2)

∑F=maN-Mgcosθ=0N=Mgcosθ

The relation between the frictional force and Normal force using the above data in equation (3) is given as:

fs=μ²Ñ²µcosθ................4

A cosine function decreases with an increase in angle θ.

Hence, the frictional force fsalso decreases with the increase θ.

05

c) Calculation of the component of the gravitational force perpendicular to the ramp

The component of the gravitational force perpendicular to the ramp using equation (2) is given by Mgcosθ.

As the cosine function decreases with increase θ.

Hence, this perpendicular component of gravitational force also decreases.

06

d) Calculation of the normal force on the box along the ramp

The normal force on the box from the ramp is defined from the part (b) calculations as,

N=Mgcosθ

The normal force also decreases with increasing anglesθ considering the cosine function.

07

e) Calculation of the maximum value of static frictional force

The maximum value of static frictional force whenθ=0° is given using equation (4) as follows:

fs,max=μ²Ñ²µ

From this, we can say that it does not depend on the angle of the inclination so it remains the same.

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!

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

Body A in Fig. 6-33 weighs 102N, and body B weighs 32N. The coefficients of friction between A and the incline are μs=0.56and μk=0.25. Angle θis 400. Let the positive direction of an xaxis be up the incline. In unit-vector notation, what is the acceleration of A if A is initially (a) at rest, (b) moving up the incline, and (c) moving down the incline?

Suppose the coefficient of static friction between the road and the tires on a car is0.60and the car has no negative lift. What speed will put the car on the verge of sliding as it rounds a level curve of30.5mradius?

Figure 6-20 shows an initially stationary block of masson a floor. A force of magnitudeis then applied at upward angleθ=20°.What is the magnitude of the acceleration of the block across the floor if the friction coefficients are (a)μs=0.600andμk=0.500and (b)μs=0.400andμk=0.300?

You testify as an expert witness in a case involving an accident in which car A slid into the rear of car B, which was stopped at a red light along a road headed down a hill (Fig. 6-25). You find that the slope of the hill is θ=12.00, that the cars were separated by distance d=24.0mv0=18.0m/swhen the driver of car A put the car into a slide (it lacked any automatic anti-brake-lock system), and that the speed of car A at the onset of braking was v0=18.0m/s.With what speed did car A hit car B if the coefficient of kinetic friction was (a) 0.60(dry road surface) and (b) 0.10(road surface covered with wet leaves)?


A person riding a Ferris wheel moves through positions at (1) the top, (2) the bottom, and (3) mid height. If the wheel rotates at a constant rate, rank these three positions according to (a) the magnitude of the person’s centripetal acceleration, (b) the magnitude of the net centripetal force on the person, and (c) the magnitude of the normal force on the person, greatest first.

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.