Physics, 25.02.2020 21:21 khaz1430

The factory worker pushes a 30.0 kg crate a distance of 3.3 mm along a level floor at a constant velocity by pushing downward at an angle of 30 ∘∘ below the horizontal. The coefficient of kinetic friction between the crate and the floor is 0.25. What magnitude of force must the worker apply to move the crate at constant velocity?

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Answer from: debbys4153

The magnitude of force is 99.2 N.

Given:

m=30.0kg

To solve the problem, we have to write the equation of the forces along two perpendicular directions.

Along the vertical direction:

$N+F Sin \theta-mg=0$

$N=mg-Fsin \theta$ ................(i)

where,

N is the normal reaction of the floor on the crate

F is the force applied by the worker

$\theta=30^o$ is the angle at which the force is applied

mg is the weight of the crate, with

m = 30.0 kg (mass of the crate)-given

$g=9.8ms^{-2}$ (acceleration of gravity)

Along Horizontal direction:

$F cos \theta-\muN=ma$

where,

$\mu N$ is the force of friction, where

$\mu=0.25$ is the coefficient of friction-given

a is the acceleration; since the crate is moving at constant velocity, the acceleration is zero,

∴ $Fcos\theta-\mu N=0$ ...............(ii)

on equating (i) and (ii)

$Fcos \theta-\mu (mg-Fsin \theta)=0\\\\F=\frac{\mu mg}{cos \theta+\mu sin \theta}\\\\F=\frac{0.25*30.0*9.8}{cos(-30^o)+0.25 *sin (-30^o)} =99.2N$

The magnitude of force is 99.2 N.

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Answer from: potatocow

99.2 N

Explanation:

To solve the problem, we have to write the equation of the forces along two perpendicular directions.

Along the vertical direction:

$N+F sin \theta -mg=0$ (1)

where

N is the normal reaction of the floor on the crate

F is the force applied by the worker

$\theta=-30^{\circ}$ is the angle at which the force is applied

mg is the weight of the crate, with

m = 30.0 kg (mass of the crate)

g=9.8 m/s^2 (acceleration of gravity)

Along the horizontal direction:

$F cos \theta - \mu N = ma$

where

$\mu N$ is the force of friction, where

$\mu=0.25$ is the coefficient of friction

and

a is the acceleration; since the crate is moving at constant velocity, the acceleration is zero, so

a=0

And the equation becomes

$Fcos \theta - \mu N =0$ (2)

From eq.(1) we get:

$N=mg-F sin \theta$

And substituting into (2) we get

$Fcos \theta -\mu(mg-F sin \theta)=0\\F cos \theta -\mu mg + \mu F sin \theta =0\\F(cos \theta+\mu sin \theta)=\mu mg\\F=\frac{\mu mg}{cos \theta + \mu sin \theta}=\frac{(0.25)(30.0)(9.8)}{cos(-30^{\circ})+(0.25)(sin(-30^{\circ}))}=99.2 N$

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