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,

Question

1 month ago

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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 [3.4K] · Answer 1 · 2026-03-01T06:06:10+03:00

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

Explanation:

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.

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.