I am pulling a 500 kilogram box, with an increasing speed, up a plane inclined at an angle of 41° with a force of 4900 Newtons. Unlike my scale, you may assume that I am massless. Furthermore, you...

I am pulling a 500 kilogram box, with an increasing speed, up a plane inclined at an angle of 41° with a force of 4900 Newtons. Unlike my scale, you may assume that I am massless. Furthermore, you may assume that the inclined plane is frictionless and that the string that I’m using to pull the box is massless as well.


a. Draw a Free Body Diagram for the box.
b. Determine the magnitude of the net acceleration of the box.
c. What is the magnitude of the normal force felt by the box?

Expert Answers
ishpiro eNotes educator| Certified Educator

a) Free-body diagram is in the attached image. Notice the three forces involved: gravity (`mvecg` ), normal force `vecN` and the force the box is pulled with, `vecF` .

b) and c) The second Newton's Law states that the net vector sum of forces equals mass times acceleration:

`mvecg + vecN + vecF = mveca`

This is a vector equation, and can be solved by resolving it into x- and y-components.

If the box is being pulled up the incline with the increasing speed, its acceleration is directed up the incline. Therefore let's choose the x-axis to be directed up the incline as well, and the y-axis perpendicular to the incline and up.

This arrangement is convenient because then acceleration has only x-component and normal force has no-x component. Gravity has both x and y component. If you draw the line perpendicular to the incline from the tip of the vector `mvecg` , you will find from the resultant right triangle that the x-component of gravity will be  `-mgsin(theta)` and the y-component of gravity will be `-mgsin(theta)` , where `theta` is the angle of the incline. (Please see the attached image.)

So the vector equation above, when projected on x-axis

`-mgsin(theta) + F = ma` . From here, the acceleration of the box can be determined:

`a = F/m - gsin(theta)` .

Plugging in the values given, we get

`a = 4900/500-9.8sin(41) = 3.37 m/s^2`

When projected on y-axis, the vector equation becomes

`-mgcos(theta) + N = 0`

From here, the normal force can be found as

`N = mgcos(theta) = 500*9.8cos(41) = 3,698 N`

The magnitude of the net acceleration is 3.37 m/s^2 and the magnitude of the normal force is 3,698 N.

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