A student decides to move a box of books into her dormitory room by pulling on a
ID: 1778698 • Letter: A
Question
A student decides to move a box of books into her dormitory room by pulling on a rope attached to the box. She pulls with a force of 73.4 N at an angle of 28.0° above the horizontal. The box has a mass of 21.0 kg, and the coefficient of kinetic friction between box and floor is 0.300. (Indicate the direction with the sign of your answer.)
(a) Find the acceleration of the box. (Assume that the +x-axis is to the right and the +y-axis is up along the page.)
magnitude in m/s^2
direction? ° counterclockwise from the +x-axis
(b) The student now starts moving the box up a 10.0° incline, keeping her 73.4 N force directed at 28.0° above the line of the incline. If the coefficient of friction is unchanged, what is the new acceleration of the box? (Assume that the +x-axis is along the incline and the +y-axis is upwards and perpendicular to the incline.)
magnitude?
direction? ° counterclockwise from the incline
magnitude?
direction? ° counterclockwise from the incline
Explanation / Answer
a) First of all, resolve the 73.4 N force into its x and y components to make solving this easier.
In the x-direction, F = (73.4 N)cos28 = 64.81 N
In the y-direction, F = (73.4 N)sin28 = 34.46 N
The forces in the y-direction are gravity, the y-component of the applied force, and the normal force. The only unknown here is the normal force since you know the force due to gravity (9.8 m/s^2 * 21 kg = 205.8 N) and the y-component of the applied force.
Net force in the y = 0
--> 34.46 N - 205.8 N + F(normal) = 0
--> F(normal) = 171.34 N
Ok, so now you have the normal force which you need to find the force of friction. Remember, F(friction) = (coefficient of friction)*(normal force)
--> F(friction) = (0.300)(171.34 N) = 51.402 N
Remember that friction always opposes the motion of an object.
Therefore, net force in the x direction = Force applied in the x direction - Force of friction
= 64.81 N - 51.402 N = 13.408 N
Therefore, the net force acting on the box is 13.408 N in the x direction.
You can now find the acceleration of the box using F = ma
--> 13.408 N = (21 kg)a
--> a = +0.638 m/s^2. (ans)
0 degree ccw from the x-axis
b) This time resolve all the forces into forces parallel to the incline and forces perpendicular to the incline. I'll call x the parallel direction and y the perpendicular (not to be confused with horizontal and vertical).
Applied force in the x direction = 64.81 N
Applied force in the y direction = 34.46 N
Force of gravity in the x = (9.81)(21 kg)(sin 10.0) = 35.77 N
Force of gravity in the y = (9.81)(21 kg)(cos 10.0) = 202.88 N
Again, find the normal force, noting that the net force perpendicular to the ramp is 0 since the box is not accelerating in that direction:
Net force y = 0
--> 34.46 N - 202.88 N + F(normal) = 0
--> F(normal) = 168.42 N
The force of friction = (168.42 N)(0.300) = 50.53 N
Ok, so now you have all the forces in the x direction.
Net force in the x = 64.81 N – 35.77 N – 50.53 N = -21.49 N
F = ma
--> -21.49 N = (21 kg)a
--> a = -1.02 m/s^2 UP the ramp.
10 degrees ccw from the x-axis
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