What is the force of friction when a body weighing 120 kg left from the top of a surface at 30 degree to the horizontal slides down at a constant velocity?

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The body weighing 120 kg is allowed to slide down from the top of a surface that makes an angle of 30 degree with the horizontal. The angle it makes with the vertical is equal to 90-30 = 60 degrees.

In the absence of friction, the body would accelerate downwards due to the gravitational force of attraction of 9.8 m/s^2 acting vertically downwards. The component of the vertical force of gravitation along the slope is equal to m*g*cos 60.

As the body moves down at a constant velocity, the force of gravitational attraction is equal to the force of friction which is in the opposite direction.

This gives the force of friction as m*g*cos 60 = 120*9.8*cos 60 = 588 N

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