Physics, 10.03.2020 08:24 limelight11

A worker exerts a pulling force on a box. The worker exerts this force by attaching a rope to the box and pulling on the rope so that the rope makes a constant angle of 36.8° above the horizontal. The coefficient of kinetic friction between the box and the floor is 0.10. It takes the worker 5.12 seconds to move the box 10 m from rest. A. What is the magnitude of the pulling force that the worker exerts on the box? Express your answer as a percentage of the magnitude of the gravitational force acting on the box. Indicate any assumptions that you made.

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Answer from: zeesharpe05

The magnitude of the pulling force is 20.66% of the gravitational force acting on the box

Explanation:

Accelerated Motion

The net force exerted on a body is the (vector) sum of all forces applied to the body. The net force can be decomposed in its rectangular components and the dynamics of the body can be studied in each direction x,y separately.

Let's start off by calculating the acceleration the worker gives to the box when pulling it. The distance traveled by the box initially at rest in a time t at an acceleration a is given by

$\displaystyle x=\frac{at^2}{2}$

Solving for a

$\displaystyle a=\frac{2x}{t^2}$

$\displaystyle a=\frac{2\cdot 10}{5.12^2}$

$a=0.763\ m/s^2$

Now we analyze the geometric of the forces applied to the box. Please refer to the free body diagram provided below.

The forces in the y-axis must be in equilibrium since no movement takes place there, thus, being g the acceleration of gravity:

T_y+N=m.g

There Ty is the vertical component of the tension of the rope, N is the normal force, and m is the mass of the box

The decomposition of T gives us

$T_y=Tsin\theta$

$T_x=Tcos\theta$

Solving the above equation for N

$Tsin\theta+N=m.g$

$N=m.g-Tsin\theta\text{..........[1]}$

Now for the x-axis, there are two forces acting on the box, the x-component of the tension and the friction force Fr. Those forces are not equilibrated, thus acceleration is produced:

$Tcos\theta-F_r=m.a$

Recalling that

$F_r=\mu N$

$Tcos\theta-\mu N=m.a$

Replacing N from [1]

$Tcos\theta-\mu (m.g-Tsin\theta)=m.a$

Operating

$Tcos\theta-\mu m.g+\mu Tsin\theta=m.a$

Solving for T

$T(cos\theta+\mu sin\theta)=m.a+\mu m.g$

$\displaystyle T=\frac{m.a+\mu m.g}{cos\theta+\mu sin\theta}$

$\displaystyle T=m\frac{a+\mu g}{cos\theta+\mu sin\theta}$

We don't know the value of m, thus we'll plug in the rest of the data

$\displaystyle T=m\frac{0.763+0.10\cdot 9.8}{cos36.8^o+0.10 sin36.8^o}$

T=2.0252m

Dividing by the weight of the box m.g

T/(m.g)=2.0252/9.8=0.2066

Thus, the magnitude of the pulling force is 20.66% of the gravitational force acting on the box

Answer from: Quest

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Answer from: Quest

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