Question

A worker drags a box of mass 50 kg across a horizontal floor. The worker attaches a rope to the box and pulls on the rope so that the rope makes a constant angle of 36.8 o 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. Find an†expression†for the magnitude of the pulling force that the worker exerts on the rope.

B. Find a†numerical†value†for the magnitude of the pulling force that the worker exerts on the rope. Express your answer as a percentage of the magnitude of the gravitational force acting on the box.

Answer 1

Answer:

A) $F=\displaystyle{\frac{m(a+ \mu g)}{\cos(\alpha) -\mu \sin(\alpha)} }$with $a = \frac{2\Delta x}{t^2}$

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:

$\sum_{y} F = 0\\F\sin(\alpha) + N - F_g = 0\\N = F_g - F\sin(\alpha)\\N = mg - F\sin(\alpha)$ (1)

where $\alpha$ is the given angle of the rope with the horizontal.

In the x-axis, the box moves with an acceleration $a$ that can be calculated as a function of the given displacement and time interval as:

$a = \frac{2\Delta x}{t^2}$ and the force equation is:

$\sum_{x} F = ma\\F\cos(\alpha) -f =ma\\F\cos(\alpha) - \mu N = ma$

now substituting $N$ from equation (1):

$F\cos(\alpha) - \mu N = ma\\F\cos(\alpha) - \mu (mg - F\sin(\alpha))=ma\\F\cos(\alpha) - \mu mg -\mu F\sin(\alpha)=ma\\F\cos(\alpha) -\mu F\sin(\alpha)=ma+ \mu mg\\F(\cos(\alpha) -\mu \sin(\alpha))=ma+ \mu mg\\\boxed{F=\frac{m(a+ \mu g)}{\cos(\alpha) -\mu \sin(\alpha)}}$

and putting the expression for the acceleration:

$\boxed{F=\frac{m \frac{2\Delta x}{t^2}+ \mu mg}{\cos(\alpha) -\mu \sin(\alpha)}}$ (2)

which is the requested expression

B) Substituting the values in (2) and using $g=9.8\ \rm{ms^{-2}}$:

$F=\frac{50 \cdot ( 0.762939+ 0.1\cdot \cdot 9.8)}{\cos(36.8^{\circ}) -0.1 \sin(36.8^{\circ})}\\F=\frac{87.147}{0.740829} \\F=11.634\ \rm{N}$

The gravitational force acting on the box is:

$F_g=mg=50\cdot9.8=490\ \rm{N}$

$F/F_g = (117.634/490)\cdot 100 = 24 \%$

Thus, the magnitude of the pulling force is 24 % of the magnitude of the gravitational force acting on the box.