Question

A block of mass m is placed at rest on a horizontal rough surface with angle of friction Φ. The block is pulled with a force F at an angle α with the horizontal . The minimum value of F required to move block is

Answer 1

This is actually a tricky one since many people will end up getting

minimum force as f=μ.mgf=μ.mg

However Just read the question again , it says minimum force (There is no mention of minimum horizontal force)

Now as we know Friction force∼Normalforce∼Normal

i.e. : if we minimize the normal reaction we can minimize the friction at the same time we also need to provide enough Horizontal force to ensure block gets a horizontal velocity

Something like shown in figure !

Now the task is to find optimal angle “x” and Magnitude F

Applying simple physics : (Assuming Normal reaction from surface is N)

Balancing Vertical force

Fsinx+N=mgFsinx+N=mg —>(1)

N=mg−FsinxN=mg−Fsinx

Balancing Horizontal force

Fcosx=FrictionFcosx=Friction

=> Fcosx=μ.NFcosx=μ.N

=> Fcosx=μ.(mg−Fsinx)Fcosx=μ.(mg−Fsinx)

=> F(cosx+μ.sinx)=μ.mgF(cosx+μ.sinx)=μ.mg

=> F=μ.mgcosx+μ.sinxF=μ.mgcosx+μ.sinx ==>(2)

Now to minimize F(2) we need to maximize denominator of 2

maxmax{cosx+μ.sinxcosx+μ.sinx}=1+(μ)2−−−−−−−√=1+(μ)2 Basic Trigo

So minimum force needed would be

F=μ.mg1+(μ)2√F=μ.mg1+(μ)2

Answer 2

Hey mate ^_^

=> The component of weight Mg of the block along the inclined plane =Mgsinθ

=> The minimum frictional force to be overcome is also Mgsinθ

=> To make the block just move up the plane the minimum force applied must overcome the component Mgsinθ of gravitational force as well as the frictional force:

Mgsinθ=2Mgsinθ

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Hence the answer is 2Mgsinθ
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