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An athlete trains by dragging a heavy load across a rough horizontal surface. The athlete exerts a force of magnitude f on the load at an angle of 25° to the horizontal. (a) once the load is moving at a steady speed, the average horizontal frictional force acting on the load is 470 n. Calculate the average value of f that will enable the load to move at constant speed. |
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Answer» The FORCE APPLIED f by the athlete under steady speed is 518.76 N Explanation: The force f applied by the user has two components which are given by, vertical component = f sinФ horizontal component = f cosФ In case of steady speed the NET horizontal force applied will be EQUAL to the FRICTION force. Given friction force = 470 n, Ф = load angle = 25° Hence, f cos 25 = 470 => f = 470/cos 25 => f = 470/0.906 => f = 518.76 n. Hence the total force applied under steady speed is 518.76 N. |
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