Finding coefficient of static friction with velocity and radius

AI Thread Summary
To determine the minimum coefficient of static friction needed to prevent a laptop from sliding on a bus seat during a turn, the centripetal force equation Fc = mv^2/r is essential. The force of static friction (Fs) must counteract the centripetal force, and the normal force (Fn) balances the weight of the laptop. The correct relationship is Fs = μ_s * Fn, where μ_s is the coefficient of static friction. The normal force can be assumed equal to the weight if the seat is horizontal, simplifying calculations. The discussion concludes that the coefficient of static friction is approximately 0.4, indicating that the calculations align with this value.
phee
Messages
36
Reaction score
0

Homework Statement



A bus passenger has her laptop sitting on the flat seat beside her as the bus, traveling at 10.0 m/s, goes around a turn with a radius of 25.0 m. What minimum coefficient of static friction is necessary to keep the laptop from sliding?

Given:

V = 10 m/s
r = 25.0 m

Homework Equations



Fc = mv ^ 2 / r

f = Us x Fn

The Attempt at a Solution



Fs = Force of static frictionSo far I have:

Fc = mv^2 / r

Fs + mg = mv^2 / r

Fs = v^2 / r - g

Fs = 10 ^ 2 / 25 - 9.8

Fs = -5.8I don't think I am on the right track with this as my Fs is a negative. Any hint to the proper direction would be much appreciated. Thank you
 
Physics news on Phys.org
I do not understand what Fs is and why do you add the weight to it. Since your object does not slide you can find a relation involving only the centripetal force and the friction force. From that you solve for the mu_s_min
 
bp_psy said:
I do not understand what Fs is and why do you add the weight to it. Since your object does not slide you can find a relation involving only the centripetal force and the friction force. From that you solve for the mu_s_min

Fs = Force of Static friction ?

Do I not need to find Fs?
 
phee said:
Fs = Force of Static friction ?

Do I not need to find Fs?

It is the adding of mg that is wrong. gravity acts vertically down. Fs acts horizontally. The centripetal force required is horizontal.

The normal force would be useful on two fronts. It balances the weight force and along with the coefficient of friction, enables you to calculate that coefficient.

Fs + mg + Fn = mv^2 / r is what you needed - and that is a vector sum formula so don't just go numerically adding and subtracting bits unless they are in the same direction
 
PeterO said:
The normal force would be useful on two fronts. It balances the weight force and along with the coefficient of friction, enables you to calculate that coefficient.

So how would I go about finding Fn and Fs ?

This is where I am totally lost, the only givens I have are the radius = 25 m and V = 10 m/s
 
phee said:
So how would I go about finding Fn and Fs ?

This is where I am totally lost, the only givens I have are the radius = 25 m and V = 10 m/s

Your 3rd equation should have been

Fs + mg + Fn = mv^2 / r

If we assume the seat is horizontal [a reasonable assumption] then the normal force will be the same size as the weight force, so they will balance, and could have been left out all together.

Fs = mju * Fn

where mju is the coefficient of friction.

Ultimately the mass will cancel out - which is useful as it would be unfortunate if only the light/heavy passengers in a bus could go around the corner with the bus.
 
I am also working on this question and stumbled upon the answer of 0.4. Without plugging in all of my work is anyone able to tell me if I am correct or completely wrong?
 
Back
Top