A 5.00 kg block is placed on top of a 12.0 kg block that rests on a frictionless

Question

A 5.00 kg block is placed on top of a 12.0 kg block that rests on a frictionless table. The coefficient of static friction between the two blocks is 0.6. What is the maximum horizontal force that can be applied before the 5.00 kg black begins to slip relative to the 12.0 kg block, if the force is applied to a. the more massive block and b. the less massive block?

Explanation / Answer

First, consider the force applied to the 5 kg block. Looking at FBDs of each and using F=m*a Find an F such that the acceleration of the 5 kg block is greater than the acceleration of the 12 kg block F-5*g*0.6=5*a1 5*g*0.6=12*a2 solve for a2 = 2.4525 plug into the first equation F>5*2.4525+5*g*0.6 F> 41.7 N Similarly when the force is applied to the 12 kg block, F such that the acceleration of the 12 kg block is greater than the acceleration of the 5 kg block F-5*g*0.6=12*a2 5*g*0.6=5*a1 a1= 9.81*0.6 F>12*9.81+5*9.81*0.6 F> 100

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