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A box mass 5 kg is placed on a surface. The coefficient of friction between surface and...
A box mass 5 kg is placed on a surface. The coefficient of friction between surface and box is 0.5. Another box is placed on this box with coefficient of friction 0.4. What should be the mass of the topmost box to so there is 0 acceleration when the lower box is pulled by force 50 N.
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A structure consists of two boxes, the lower one has mass 5 kg and the coefficient of friction with the ground is 0.5. There is another box placed on this box with mass M. The coefficient of friction between the two boxes is 0.4. The mass M has to be determined so that when the lower box is pulled with a force 50 N, it does not accelerate.
The force of friction between the lower box and the ground is equal to (5 + M)*g*0.5 where M is the mass of the upper box. If there is no acceleration when a force of 50 N is applied, the net force on the box is 0.
This is true when (5 + M)*g*0.5 = 50
=> 5 + M = 10.2
=> M = 5.2
The required mass of the box placed on top is 5.2 kg
Posted by justaguide on September 21, 2013 at 3:29 PM (Answer #1)
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