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

Update:

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

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  • 7 months ago
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    ...

    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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    • janina53097 months agoReport

      Thanks! Very easy to understand.

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  • 7 months ago

    equals weight x coefficient of static friction.

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