Question

a mass of 100kg is resting on a rough inclined plane of an angle 30 degree if the coefficient of friction is 1 by root3 find the greatest and the least force that acting parallel to the plane in the both cases just maintain the mass is equilibrium
if you are the student of class 11 science than only. give me answer

Answer 1

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Answer 2

Answer:

100kgf,zero .

Solution :

Here, m=100kg.θ=30∘,μ=13–√

Force acting on the block down the plane

=mgsinθ=100gsin30∘=50gN=50kgf

Force of friction =f=μR=μmgcosθ

=13–√mgcos30∘=13–√100g3–√2=50gN

=50kgf

As the force of friction just balances the force acting on the block down the plane therefore mass will be in equilibrium The least force required is zero

When we tend to move the mass up the inclined plane force of friction acts down the plane Therefore greatest force required just to start upward motion (or to maintain the block in equilibrium ) =mgsinθ+f

50kgf+50kgf=100kgf