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
Answers
Answered by
10
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
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
Attachments:
Answered by
7
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
==========================
Hence the answer is 2Mgsinθ
==========================
=> 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
==========================
Hence the answer is 2Mgsinθ
==========================
Similar questions