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?

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ishpiro | College Teacher | (Level 1) Educator

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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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