The surfaces between a 5 kg block, the 22 kg wedge and between the 22 kg wedge a

Question

The surfaces between a 5 kg block, the 22 kg wedge and between the 22 kg wedge and the horizontal plane are smooth (without fric- tion).

The acceleration of gravity is 9.8 m/s2 .

A block is released on the inclined plane (top side of the wedge).

What is the force F which must be exerted on the 22 kg block in order that the 5 kg block does not move up or down the plane?

Answer in units of N

The surfaces between a 5 kg block, the 22 kg wedge and between the 22 kg wedge and the horizontal plane are smooth (without friction). The acceleration of gravity is 9.8 m/s2 . A block is released on the inclined plane (top side of the wedge). What is the force F which must be exerted on the 22 kg block in order that the 5 kg block does not move up or down the plane? Answer in units of N

Explanation / Answer

Nsin19 = 5a

Ncos19 = 5g

tan19 = a/g

a = gtan19

F = (22 +5)a = 27gtan19 = 91.11 N

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