Respuesta :
Answer:
A) [tex]F=\displaystyle{\frac{m(a+ \mu g)}{\cos(\alpha) -\mu \sin(\alpha)} }[/tex]with [tex]a = \frac{2\Delta x}{t^2}[/tex]
B) The magnitude of the pulling force is 24 % of the magnitude of the gravitational force acting on the box
Step-by-step explanation:
A) The forces acting on the box are the pulling force that the worker exerts on the rope, the friction, the normal force, and the gravitational force. See the attached diagram. Since the pulling force on the rope makes an angle with the horizontal, this will have components in the x and y-axis (see diagram).
In the y-axis, the box does not move, therefore:
[tex]\sum_{y} F = 0\\F\sin(\alpha) + N - F_g = 0\\N = F_g - F\sin(\alpha)\\N = mg - F\sin(\alpha)[/tex] (1)
where [tex]\alpha[/tex] is the given angle of the rope with the horizontal.
In the x-axis, the box moves with an acceleration [tex]a[/tex] that can be calculated as a function of the given displacement and time interval as:
[tex]a = \frac{2\Delta x}{t^2}[/tex] and the force equation is:
[tex]\sum_{x} F = ma\\F\cos(\alpha) -f =ma\\F\cos(\alpha) - \mu N = ma[/tex]
now substituting [tex]N[/tex] from equation (1):
[tex]F\cos(\alpha) - \mu N = ma\\F\cos(\alpha) - \mu (mg - F\sin(\alpha))=ma\\F\cos(\alpha) - \mu mg -\mu F\sin(\alpha)=ma\\F\cos(\alpha) -\mu F\sin(\alpha)=ma+ \mu mg\\F(\cos(\alpha) -\mu \sin(\alpha))=ma+ \mu mg\\\boxed{F=\frac{m(a+ \mu g)}{\cos(\alpha) -\mu \sin(\alpha)}}[/tex]
and putting the expression for the acceleration:
[tex]\boxed{F=\frac{m \frac{2\Delta x}{t^2}+ \mu mg}{\cos(\alpha) -\mu \sin(\alpha)}}[/tex] (2)
which is the requested expression
B) Substituting the values in (2) and using [tex]g=9.8\ \rm{ms^{-2}}[/tex]:
[tex]F=\frac{50 \cdot ( 0.762939+ 0.1\cdot \cdot 9.8)}{\cos(36.8^{\circ}) -0.1 \sin(36.8^{\circ})}\\F=\frac{87.147}{0.740829} \\F=11.634\ \rm{N}[/tex]
The gravitational force acting on the box is:
[tex]F_g=mg=50\cdot9.8=490\ \rm{N}[/tex]
[tex]F/F_g = (117.634/490)\cdot 100 = 24 \%[/tex]
Thus, the magnitude of the pulling force is 24 % of the magnitude of the gravitational force acting on the box.
Friction force is the force which resist the pulling force on solid surface.
- A) The expression for the magnitude of the pulling force that the worker exerts on the rope is.
[tex]F=\dfrac{m(\dfrac{2\Delta x +\mu gt)}{t} }{\cos\theta-\mu\sin\theta}[/tex]
- B)the magnitude of the pulling force that the worker exerts on the rope in percentage of gravitational force is 24%.
Given information-
The mass of the box is 50 kg.
The angle made by the rope is 36.8 degrees above the horizontal.
The coefficient of kinetic friction between the box and the floor is 0.10.
The total time taken to box 10 m is 5.12 seconds.
What is friction force?
Friction force is the force which resist the pulling force on solid surface.
- A) Expression for the magnitude of the pulling force that the worker exerts on the rope-
Net force of the box for the y axis is equal to zero, as box does not move in the y axis thus,
[tex]\sum y=0[/tex]
The normal force acting on the box is equal to the difference of the gravitation force and pulling force. Thus,
[tex]N=mg-Fsin\theta[/tex]
Net force of the box for the x-axis is equal to displacement of the box.
It can be given as,
[tex]\sum x=m\dfrac{\Delta x}{t}[/tex]
Here x is the displacement of the box.
The force acting on the x axis is the difference of the pulling force to the friction force. Thus,
[tex]F\cos\theta-\mu N=m\dfrac{\Delta x}{t} \\N= \dfrac{F\cos\theta-m\dfrac{\Delta x}{t}}{\mu}[/tex]
Here [tex]\mu[/tex] is the friction coefficient.
Compare the two value of normal force,
[tex]mg-Fsin\theta= \dfrac{F\cos\theta-m\dfrac{\Delta x}{t}}{\mu}[/tex]
Solve for [tex]F[/tex],
[tex]F=\dfrac{m(\dfrac{2\Delta x +\mu gt)}{t} }{\cos\theta-\mu\sin\theta}[/tex]
Hence the expression for the magnitude of the pulling force that the worker exerts on the rope is.
[tex]F=\dfrac{m(\dfrac{2\Delta x +\mu gt)}{t} }{\cos\theta-\mu\sin\theta}[/tex]
- B) Magnitude of the pulling force that the worker exerts on the rope-
Put the values in the above obtained formula,
[tex]F=\dfrac{50(\dfrac{2\times 10 +0.10\times9.81\times5.12)}{5.12} }{\cos\ 36.8-0.10\times\sin36.8}[/tex]
[tex]F=11.635[/tex]
The gravitational force is,
[tex]F_g=mg\\F_g=50\times9.81\\F_g=490[/tex]
Hence the difference in percentage can be given as,
[tex]p=\dfrac{11.635}{490} \times100\\p=24[/tex]
Hence the magnitude of the pulling force that the worker exerts on the rope in percentage of gravitational force is 24%.
Hence,
- A) The expression for the magnitude of the pulling force that the worker exerts on the rope is.
[tex]F=\dfrac{m(\dfrac{2\Delta x +\mu gt)}{t} }{\cos\theta-\mu\sin\theta}[/tex]
- B)the magnitude of the pulling force that the worker exerts on the rope in percentage of gravitational force is 24%.
Learn more about the friction force here;
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