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4 days ago

15

A 60.0-kg crate rests on a level floor at a shipping dock. The coefficients of static and kinetic friction are 0.760 and 0.410,

respectively. What horizontal pushing force is required to (a) just start the crate moving and (b) slide the crate across the dock at constant speed

1 answer:

ValentinkaMS [1.1K]4 days ago

6 0

The initial horizontal force needed to start moving the crate is 447 N.

The constant horizontal force required to push the crate across the dock at a steady pace is 241 N.

Explanation:

According to the coefficient of static friction, the following holds:

μ $_{s}$ = $\frac{F_{appl} }{W}= \frac{F_{s} }{N}$ ,

where $F_{appl}$ represents the horizontal force applied,

W = mg signifies the weight of the crate acting downwards,

$F_{s}$ denotes the static friction force, which opposes the horizontal force and is equal to it,

N is the reaction force acting upward that is equal to the weight of the crate.

Using this formula, we can determine the horizontal force needed to initiate movement of the crate:

$F_{appl} = F_{s} = u_{s}N = u_{s}mg$

= 0.760 $\times$ 60 kg $\times$ 9.8 m / s^2

= 447 N.

By the definition of coefficient of kinetic friction, we have:

u $_{k} = \frac{F_{appl} }{W} = \frac{F_{k} }{N}$ ,

where $F_{appl}$ indicates the horizontal applied force,

W = mg represents the weight of the crate directed downwards,

$F_{k}$ is the kinetic friction force opposing the horizontal force and equal to it,

N represents the reaction force directed upwards and equal to the weight of the crate.

Thus, we find the horizontal force needed to maintain a constant speed while sliding the crate:

$F_{appl} = F_{k} = u_{k}N = u_{k}mg$

= 0.410 $\times$ 60 $\times$ 9.8

= 241 N.

The initial horizontal force required to simply initiate movement of the crate is 447 N.

The force needed to continuously slide the crate across the dock at a consistent pace is 241 N.

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