A block is moving on an inclined plane making an angle 45 degree with the horizontal and the coefficient of friction is mew. the force required to jsut push it up the inclined plane is 3 times the force required to just prevent it from sliding down. if we define N equal to 10mew then N is
Asked by tluser4
18th September 2018,
5:21 PM
Answered by Expert
Answer:
figure shows the different forces acting on the block, when it is prevented from falling or pushed up.
when the applied force Fh is ment for preventing the block from falling, friction force act against movement,
hence friction force acts upward and it is against the direction of pulling force mg×sin45.
Fh = mg×sin45 - μR = mg×sin45 - μ×mg×cos45 = ( mg/√2 )×(1-μ) .................(1)
where R is the normal reaction force.
when the applied force Fu is ment for pushing the block up, friction force act against movement,
hence friction force acts downward and it is in the same direction of pulling force mg×sin45.
Fu = mg×sin45 + μR = mg×sin45 + μ×mg×cos45 = ( mg/√2 )×(1+μ) .................(2)
It is given that, Fu = 3×Fh .............(3)
From (1), (2) and (3) we can write, ( mg/√2 )×(1+μ) = 3×( mg/√2 )×(1-μ) or μ = 0.5
N = 10×μ = 10×0.5 = 5
Answered by Expert
19th September 2018,
10:08 AM
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