# What is the magnitude of the force of static friction fs on the block?

A mass m1 = 1.0 kg, sits on a plane inclined at angle α = 30 degrees to the horizontal. μs = 0.40, while μk = 0.15. The mass m1 is attached by a massless string over a massless frictionless pulley to a bucket with mass m2 = 0.24 kg which hangs freely.

What is the magnitude of the force of static friction fs on the block?

The normal force is 8.5N, but the answer for force of static friction is 2.6N. Can someone please tell me how to work through this problem? I thought Fs was μs*N

### 2 Answers

- Old Science GuyLv 712 months agoFavourite answer
...

friction is a reaction force to whatever force is trying to cause motion

and mu Fn is a maximum possible value -

Fs variable depending on circumstances

when this block is stationary the forces are in equilibrium

force down plane = friction + weight of hanging object

(1 kg*9.81N/kg) sin30 = Fs + (0.24kg*9.81N/kg)

Fs = 2.5506 N which rounds to 2.6 N

max Fs = 0.40(9.81cos30) = 3.4

so 2.6 N is less than max therefore an acceptable value

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- billrussell42Lv 712 months ago
equals weight x coefficient of static friction.

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Thanks! Very easy to understand.