Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

In three experiments, three different horizontal forces are applied to the same block lying on the same countertop. The force magnitudes areF1=12N,F2=8N, F3=4N. In each experiment, the block remains stationary in spite of the applied force. Rank the forces according to (a) the magnitude of the static frictional force on the block from the countertop and (b) the maximum value role="math" localid="1660904123305" fs,maxof that force, greatest first.

Short Answer

Expert verified

a) The magnitude of the static frictional force on the block from the countertop is given by fs1>fs2>fs3.

b) The maximum values fsmaxof those forces are given by fs1max=fs2max=fs3max.

Step by step solution

01

The given data

The magnitudes of the force are: F1=12N, F2=8N, and F3=4N.

02

Understanding the concept of the force

We have to use Newton’s 2nd law of motion along with x and y directions on each block. Now, using the given data of the magnitudes of the force, we can get the behavior of the frictional force and the maximum frictional force.

Formulae:

The force according to Newton’s second law,

F=ma (1)

03

a) Calculation of the magnitude of the static frictional force

Free body diagram of blocks:

Blocks are stationary therefore the acceleration of the blocks along x is 0.

By using Newton’s 2nd law along the x-direction, the net of the horizontal force for the first block can be given using equation (1) as:

(Case-1)

Fx=maxF1-fs1=0fs1=F1fs1=12N

By using Newton’s 2nd law along the x-direction, the net of the horizontal force for the first block can be given using equation (1) as:

(Case-2)

Fx=maxF2=fs2=0fs2=F2fs2=8N

By using Newton’s 2nd law along the x-direction, the net of the horizontal force for the third block can be given using equation (1) as:

(Case-3)

Fx=maxF3=fs3=0fs3=F3fs3=4N

So, the block with F1=12Nforce has the greatest frictional force fs1.

Hence, the magnitude of the forces is ordered as fs1>fs2>fs3.

04

b) Calculation of the magnitude of the maximum force

The value of the magnitude fsmaxof the static frictional force is defined as,

fsmax=μN

So, fsmaxis directly proportional to the normal force N.

By using Newton’s 2nd law along the vertical direction, the net vertical forces acting on the body can be given using equation (1) as (since blocks are stationary ay=0.)

Fy=mayN-Mg=0N=Mg

Here, the mass of the blocks (M) is the same, and g is constant so normal force also remains the same in all blocks.

Therefore, fsmaxis the same for all blocks that imply the magnitude values as fs1max=fs2max=fs3max.

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 Vaia!

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

A box of canned goods slides down a ramp from street level into the basement of a grocery store with acceleration0.75m/s2directed down the ramp. The ramp makes an angle of40°with the horizontal. What is the coefficient of kinetic friction between the box and the ramp?

Figure 6-53 shows a conical pendulum, in which the bob (the small object at the lower end of the cord) moves in a horizontal circle at constant speed. (The cord sweeps out a cone as the bob rotates.) The bob has a mass of 0.040 kg, the string has length L=0.90 mand negligible mass, and the bob follows a circular path of circumference 0.94 m. What are

(a) the tension in the string and

(b) the period of the motion?

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?

In Fig. 6-58, force is applied to a crate of mass mon a floor where the coefficient of static friction between crate and floor is μs. Angle u is initially 0°but is gradually increased so that the force vector rotates clockwise in the figure. During the rotation, the magnitude Fof the force is continuously adjusted so that the crate is always on the verge of sliding. For μs=0.70, (a) plot the ratio F/mgversus θand (b) determine the angle θinfat which the ratio approaches an infinite value. (c) Does lubricating the floor increase or decrease θinf, or is the value unchanged? (d) What is θinffor μs=0.60?

In Fig. 6-57, a stuntman drives a car (without negative lift) over the top of a hill, the cross section of which can be approximated by a circle of radius R = 250 m. What is the greatest speed at which he can drive without the car leaving the road at the top of 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.

Sign-up for free