Respuesta :
Answer:
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:
[tex]N+F sin \theta -mg=0[/tex] (1)
where
N is the normal reaction of the floor on the crate
F is the force applied by the worker
[tex]\theta=-30^{\circ}[/tex] 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)
[tex]g=9.8 m/s^2[/tex] (acceleration of gravity)
Along the horizontal direction:
[tex]F cos \theta - \mu N = ma[/tex]
where
[tex]\mu N[/tex] is the force of friction, where
[tex]\mu=0.25[/tex] is the coefficient of friction
and
a is the acceleration; since the crate is moving at constant velocity, the acceleration is zero, so
[tex]a=0[/tex]
And the equation becomes
[tex]Fcos \theta - \mu N =0[/tex] (2)
From eq.(1) we get:
[tex]N=mg-F sin \theta[/tex]
And substituting into (2) we get
[tex]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[/tex]
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:
[tex]N+F Sin \theta-mg=0[/tex]
OR
[tex]N=mg-Fsin \theta[/tex]................(i)
where,
N is the normal reaction of the floor on the crate
F is the force applied by the worker
[tex]\theta=30^o[/tex] 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
[tex]g=9.8ms^{-2}[/tex](acceleration of gravity)
- Along Horizontal direction:
[tex]F cos \theta-\muN=ma[/tex]
where,
[tex]\mu N[/tex] is the force of friction, where
[tex]\mu=0.25[/tex] is the coefficient of friction-given
a is the acceleration; since the crate is moving at constant velocity, the acceleration is zero,
∴[tex]Fcos\theta-\mu N=0[/tex]...............(ii)
on equating (i) and (ii)
[tex]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[/tex]
The magnitude of force is 99.2 N.
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