Suppose you are standing on a train accelerating at 0.20g. What minimum coefficient of static friction must exist between your feet and the floor if you are not to slide?

Short Answer

Expert verified

The coefficient of static friction is 0.20 between your feet and the floor.

Step by step solution

01

Step 1. Given data and assumption

The non-sliding condition is the static friction force, which is equal to the reaction force due to acceleration.

Given data:

The acceleration of the train is a=0.20g.

Assumption:

Let μsbe the coefficient of static friction between your feet and the floor.

02

Step 2. Calculation of the coefficient of static friction

The normal force acting on you is N=mg.

Then, the static friction force is

fs=μsN=μsmg

You do not slide, which means that the static friction force equals the force due to the train’s acceleration. So,

fs=maμsmg=maμs=ag

Now, substituting the values in the above equation,

μs=0.20gg=0.20

Hence, the coefficient of static friction is 0.20.

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 75 kg petty thief wants to escape from a third-story jail window. Unfortunately, a makeshift rope made of sheets tied together can support a mass of only 58 kg. How might the thief use this ‘rope’ to escape? Give a quantitative answer.

Suppose an object is accelerated by a force of 100 N. Suddenly, a second force of 100 N in the opposite direction is exerted on the object so that the forces cancel. The object

(a) is brought to rest rapidly.

(b) decelerates gradually to rest.

(c) continues at the velocity it had before the second force was applied.

(d) is brought to rest, and then it accelerates in the direction of the second force.

Estimate the average force exerted by a short-putter on a 7.0 kg shot if the shot is moved through a distance of 2.8 m and is released with a speed of 13 m/s.

A person pushes a 14.0-kg lawn mower at constant speed with a force ofF=88.0Ndirected along the handle, which is at an angle of 45.0° to the horizontal (Fig. 4–58). (a) Draw the free-body diagram showing all forces acting on the mower. Calculate (b) the horizontal friction force on the mower, then (c) the normal force exerted vertically upward on the mower by the ground. (d) What force must the person exert on the lawn mower to accelerate it from rest to 1.5 m/s in 2.5 seconds, assuming the same friction force?

FIGURE 4-58 Problem 50.

A 14.0 kg bucket is lowered vertically by a rope in which there is 163 N tension at a given instant. What is the acceleration of the bucket? Is it upward or downward?

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