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The ball of mass m is moving horizontally within the horizontal trough as shown in the figure.
The weight of the ball is equal to m*g and acts in a vertically downwards direction. The weight of the ball can be split into components that are at right angles to the surfaces inclined at 60 degrees to the horizontal and parallel to them. The component that is at perpendicular to the surface is equal to the normal force. This is given by `m*g*cos 60 = (m*g)/2` . As the coefficient of friction is `mu` , the retardation of the ball is `2*(m*g)/2*mu = mu*m*g`
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