A 2kg block is at rest on a horizontal surface wich its coefficient of static fr

Question

A 2kg block is at rest on a horizontal surface wich its coefficient of static friction is 0.3. Determine the minimum force F that must be applied to the block at an angle of 53 degrees below horizontal in order to move the block. Thank you very much. A 2kg block is at rest on a horizontal surface wich its coefficient of static friction is 0.3. Determine the minimum force F that must be applied to the block at an angle of 53 degrees below horizontal in order to move the block. Thank you very much.

Explanation / Answer

mass m = 2 kg coefficient of static friction u = 0.3 the minimum force F that must be applied to the block F = ? angle = 53 degrees Normal force N = mg - F sin 53 frcitional force f = uN = umg -uFsin53 = 5.88 - 0.2395 F If the frictional force f is equal to horizontal component of force F cos53 then it ready to move. So, F cos 53 = f 0.6018 F =5.88-0.2395 F 0.8413 F = 5.88 F = 6.989 newton

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