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Greeley
1 month ago
15

Assume that the cart is free to roll without friction and that the coefficient of static friction between the block and the cart

is μs. Derive an expression for the minimum horizontal force that must be applied to the block in order to keep it from falling to the ground. Express your answer in terms of acceleration of gravity g and some or all of the variables m, M, and μs.

Physics
2 answers:
Keith_Richards [3.2K]1 month ago
6 0

Answer:F=\frac{(M+m)g}{\mu _s}

Explanation:

Provided:

The trolley, with mass M, is allowed to roll freely without friction.

The coefficient of friction between the trolley and mass m is \mu _s.

A force F is applied to mass m.

The acceleration of the system is

a=\frac{F}{M+m}

The frictional force will counterbalance the weight of the block.

The frictional force is =\mu _sN

N=ma

\mu _sN=mg

\mu _sma=mg

\mu _s=\frac{g}{a}

F=\frac{(M+m)g}{\mu _s}

Maru [3.3K]1 month ago
3 0

Answer: N = fa/g

Explanation:

For the block not to fall, the frictional force, f, must equal the weight of the block:

f = mg. However, the block's horizontal motion is described by N = ma

Consequently,

f / N  = g/ a .

Thus, a = g/f / N

As the maximum value of f / N  is μs, we can infer that a ≥ g/μs, so that the block does not fall. Q.E.D.

N = fa/g

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