A 10 kg block is placed on top of a 30 kg block. A force of 364N is applied to t

Question

A 10 kg block is placed on top of a 30 kg block. A force of 364N is applied to the right on the lower block, and the upper block slips on the lower block (accelerating less than the lower block). The coefficient of kinetic friction between the upper block and the lower block is 0.2, and the coefficient of kinetic friction between the lower block and the floor is 0.46. How big would the coefficient of static friction between the upper and lower block have to be so that the upper block would not slip on the lower block?

Explanation / Answer

Let’s determine each friction force.

For the 10 kg block, Ff = 0.2 * 10 * 9.8 = 19.6 N
For the floor, Ff = 0.46 * (10 + 30) * 9.8 = 180.32 N

The net force on both blocks is equal to the applied force minus this friction force between the bottom block and the floor.

Net force = 364 – 180.32 = 183.68 N

183.68 = 40 * a
a = 183.68 ÷ 40 = 4.592 m/s^2

This is approximately 3.87 m/s^2. This is the acceleration of both blocks. Let’s multiply the mass of the top block by the acceleration.

F = 10 * 4.592 = 45.92N

This is approximately 45.92 N. Since this is greater than the friction force between the top and bottom block, the top block will slide off of the bottom block.

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