A block is on an inclined plane. There is a horizontal force on the block. The block is at rest. There is a certain coefficient of friction.
a) find force (magnitude and direction)
b) find normal force exerted on the block by the plane
Given: you can give the block a mass of 5.00 kg an angle of 60 degrees and the coefficient of friction 0.300
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Please see attachment.
Here is solution to this problem:
Let the inclined plane make an angle of 60 degrees with the horizontal.
let the value of g be 10 m/s^2.
Weight of the block acting vertically downward = mg = 5*10 = 50 N
Component of mg(50N) parallel and down the inclined plane = mg sin 60
Component of mg(50N) normal to the inclined plane(pressing the plane) = mg cos 60
Let the horizontal force F be applied to the ...
Resove gravitational force acting on the block kept on the inclined plane in two components; one parallel to the plane and the other perpendicular to the plane.