Quiz 2 Complete solution of problem 1

1a) The direction of the force, W due to the load is straight down. The force perpendicular to the ramp, F_perp, presses down on the ramp, while the force along the ramp, F_r is the component of the force that moves the object along the ramp. So we draw the crossection of the ramp, and we draw F_r along the ramp pointing down, and we draw F_perp perpendicular to F_r, pointing down and to the right.

We know that

.

We show this by making W the diagonal of the rectangle with sides F_perp and F_r. The finished picture looks like:



1b) The magnitude of the weight vector is the
(mass of load)(9.8m/sec²)= 490 nt.,
the direction of the weight vector is straight down, so the answer is:

-490nt

1c) We know that the points (0,0) and (10,4) are on the ramp, so the vector with tip at (0,0) and tail at (10,4) must lie on the ramp.

We can write this vector in terms of

and

by taking the difference of the x and y coordinates. If we do this we get:




1d) To get the component of the Force due to the weight in the direction of the ramp, we project the weight vector on a vector in the direction of the ramp. Let's put together what we know:

	Weight vector
	vector along ramp

The formula for the projection of W on R tells us that:

So,

1e) To get the component of the force due to the weight in the direction perpendicular to the ramp, just subtract the force along the ramp from the original force vector. This works because you

know that

.

So,




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