# Force and the Inclined Plane

Posted by Professor Cram in Weblog

Dear Professor Cram:

I'm not really sure how to tackle a problem like this… An inclined plane rests at a 37 degree angle with a box weighing 50N on it as a vector down. I'm supposed to find components of the force parallel and perpendicular to the plane. What I've come up with is to find the sides of an imaginary triangle using the force vector vs the angle (x = 50(tan(37)) from which I can find the hypotenuse, but I'm not really sure where to go from here. I could really use some pointers on this one

Jeff S.

Thank you for your excellent physics question. The inclined plane has been the bane of many a physics student, but I'll try to help you see your way through this problem.

You had a good start but fumbled a bit. (Sorry, watched alot of football this past weekend…) The inclined plane is a triangle with an acute angle of 37° and a vertical side of 50N.

The question is asking for the components of force parallel and perpendicular to the plane; because the object is being pushed up the inclined plane, the 50N force down must be counterbalanced by an equal force upwards. Thus, the perpendicular force is 50N (up).

For the parallel force, you are correct in wanting to use the tangent to calculate a side of the triangle. Remember how the tangent of an angle is the opposite side over the adjacent side? Well, the opposite side to the 37° angle is 50N, while the adjacent side is the parallel force we're looking for. Solving the tangent equation for the adjacent side:

- tan 37° = 50N/Adjacent, or
- Adjacent = 50N/(tan 37°), or
- Adjacent = 66.35N

I hope this helps. Let us know if you need anything else.

Good studying,

Professor Cram

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