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the force required to just move a body up the inclined plane is double the force required to just prevent the body form sliding down the plane. the coefficent of friction is µ . if θ is the angle of inclination of the plane the tan θ is equal to : a. µ b. 3 µ c. 2 µ d. 0.5 µ
the force required to just move a body up the inclined plane is double the force required to just prevent the body form sliding down the plane. the coefficent of friction is µ . if θ is the angle of inclination of the plane the tan θ is equal to :
a. µ
b. 3 µ
c. 2 µ
d. 0.5 µ
Rohan Bhatia force required to prevent the body from sliding θ F=mgsinθ-µmgcosθ force required to just move the body up F1=mgsinθ+µmgcosθ it is given F1=2f mgsinθ+µmgcosθ=2(mgsinθ-µmgcosθ) 0r 3µmgcosθ=mgsinθ tanθ=3µ Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly.We are all IITians and here to help you in your IIT JEE preparation. All the best Rohan. Regards, Askiitians Experts Badiuddin
force required to prevent the body from sliding
F=mgsinθ-µmgcosθ
force required to just move the body up
F1=mgsinθ+µmgcosθ
it is given
F1=2f
mgsinθ+µmgcosθ=2(mgsinθ-µmgcosθ)
0r 3µmgcosθ=mgsinθ
tanθ=3µ
In the equation for force required to prevent the body from sliding - frictional force should have been opposite to the force applied? Isn`t it? Why is it in the direction of force. Pls explain
For a sliding body, it means (mgsin t) is greater than the frictional force, so the to balance it up(to make it static) you need to add up to the frictional force to make equal to the mgsin t( the resolution of the weight of the body according to the plane).So we have F +Fr = mgsin tBut It will be F= mgsin t + Fr ( for the force to move it up the plane)
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