Okay I have no idea how to even really approach this word problem: A sled on an inclined plane weighs 500 lbs, and the plane makes an angle of 50 degrees with the horizontal. What force, perpendicular to the plane is exerted on the plane by the sled? Would anyone be willing to explain how to set up and solve????

Question

mathstudent55 · Answer

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mathstudent55 · Answer

The figure above shows the problem.
The sled is that rectangular object on the inclined plane.

mathstudent55 · Answer

The weight of the sled is represented by the vector pointing down.
The magnitude of the vector is 500 lb.

mathstudent55 · Answer

You are looking for the component of the weight vector that is in the direction perpendicular to the inclined plane.

mathstudent55 · Answer

The weight vector can be separated into two perpendicular vectors, one perpendicular to the plane and one parallel to the plane. The sum of these two vectors is the weight vector. These two vectors are perpendicular components of the weight vectors. What we are trying to do is to resolve the weight vector into its components.

mathstudent55 · Answer

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mathstudent55 · Answer

Now you need geometry (and trig) to help you.

mathstudent55 · Answer

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mathstudent55 · Answer

$ \sin 50^o = \dfrac{W_{||}}{W} $

$W_{||} = W \sin 50^o$

$\cos 50^o = \dfrac{W_{\perp}}{W} $

$W_\perp = W \cos 50^o$

We are only interested in $W_\perp$, and we know that W = 500 lb, so

$W_{\perp} = 500 \cos 50^o~lb$