Question

A crate with mass m = 4.25 kg rests on the top of a counter. The coefficient of static friction between the crate and the counter is μs = 0.62 and the coefficient of kinetic friction is μk = 0.47.
(a) Write an expression for Fm the minimum force required to produce movement of the book on the top of the bar.
(b) Solve numerically for the magnitude of the force Fm in Newtons.

Answer 1

Answer:

Explanation:

Given

mass of crate $m=4.25\ kg$

coefficient of static friction $\mu_s=0.62$

coefficient of kinetic friction $\mu_k=0.47$

For movement of book we need to provide a force which can overcome the static friction i.e. $F_m=\mu_sN$

where N=normal Reaction

and Normal reaction is equal to the weight of body

$F_m=\mu_smg$

$F_m=0.62\times 4.25\times 9.8=25.82\ N$

Answer 2

a. The  expression for Fm the minimum force required should be $\mu_smg$

b. The Fm should be 25.82 N.

Calculation of an expression for fm and value of fm:

(a)

Since A crate with mass m = 4.25 kg, the coefficient of static friction between the crate and the counter is μs = 0.62 and the coefficient of kinetic friction is μk = 0.47.

So, here an expression should be $\mu_smg$

(b)

The value of fm should be

$= 0.62 \times 4.25 \times 9.8$

= 25.82 N

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