What should the force of friction be to keep a 10 kg block at rest on a plane that makes an angle of 45 degree with the horizontal?

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justaguide | College Teacher | (Level 2) Distinguished Educator

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The force of friction is given as Fc*N, where N is the normal force and Fc is the coefficient of static friction. There is a force of gravity acting on the 10-kg block and it pulls the block vertically downwards.

We can divide this force into two components. One parallel to the plane it is placed on and the other perpendicular to the plane. The parallel component makes the block slide down and the perpendicular component acts as the normal force.

It is given that the plane is inclined at 45 degrees. cos 45 = sin 45 = 1/sqrt 2. This gives the two components as 10*g*(1/sqrt 2).

To prevent the block from sliding down the frictional force should be equal to 10*g*(1/sqrt 2) = 69.29 N (approximately)

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