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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.

Respuesta :

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

Explanation:

Given

Trolley of mass M is free to roll without friction

coefficient of friction between trolley and mass m is [tex]\mu _s[/tex]

Force F is applied on mass m

Acceleration of the system is

[tex]a=\frac{F}{M+m}[/tex]

friction Force will balance weight of block

friction force[tex]=\mu _sN[/tex]

[tex]N=ma[/tex]

[tex]\mu _sN=mg[/tex]

[tex]\mu _sma=mg[/tex]

[tex]\mu _s=\frac{g}{a}[/tex]

[tex]F=\frac{(M+m)g}{\mu _s}[/tex]

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oyesobas912

Answer: N = fa/g

Explanation:

if the  block is not to fall,the friction force, f , must balance the block's weight :

f = mg. But the horizontal motion of the block is given by N = ma

Therefore,

f / N  = g/ a

or a = g/f / N

Since the maximum value of f / N  is μs, we must have a ≥ g/μs, if the block is not to fall. Q.E.D

N = fa/g

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