Question

a box with the weight of 50N rests on a horizontal surface. a person pulls horizontally on it with a force of 10N and it does not move. to start it moving a second person pulls vertically upward on the box. if the coefficient of static friction is 0.4 what is the smallest vertical force for which the box moves?

Answer 1

Answer on Question #44437, Physics, Mechanics | Kinematics | Dynamics

A box with the weight of $50\mathrm{N}$ rests on a horizontal surface. A person pulls horizontally on it with a force of $10\mathrm{N}$ and it does not move. To start it moving a second person pulls vertically upward on the box. If the coefficient of static friction is 0.4 what is the smallest vertical force for which the box moves?

Solution:

Given:

$W = 50\mathrm{N}$

$F_{h} = 10\mathrm{N}$

$\mu = 0.4$

$F_{v} = ?$

Draw a FBD

The normal force acting on the box is

$N = W - F_{v}$

When the box starts moving, the friction force is equal to $F_{h}$

$f = F_{h}$

But, friction force is

$f = \mu N = \mu (W - F_{v})$

Thus,

$\mu (W - F_{v}) = F_{h}$

$F_{v} = W - \frac{F_{h}}{\mu} = 50 - \frac{10}{0.4} = 25\mathrm{N}$

Answer: $F_{v} = 25 \mathrm{~N}$ .

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