# Crank Force

I'm trying to establish the force in the horizontal direction of a crank when the force that is tangent to the crank is 10N.

Zero degrees is on top of a circle. Clearly, when the crank is vertical which corresponds to it being at zero degrees, the force in the horizontal plane is normal to the crank and is 10N. When the crank is at 90 degrees, the force is vertical and zero in the horizontal.

Now, when the crank is at 45 degrees, the force is acting partly vertical, partly horizontally in equal values, so the the horizontal force should be 5N.

I was hoping to apply either sine, cosine or tangent math to tell me the horizontal force, but neither do. Sin 45 is 0.707, cos 45 is 0.707 and tan 45 is 1. That surprises me that non of the trig. math I'm familiar with has found application in this problem.

## Answers and Replies

Nugatory
Mentor
I was hoping to apply either sine, cosine or tangent math to tell me the horizontal force, but neither do.
You are right that ##\cos{45}=\sqrt{2}/2\approx.707##. How is is that not working?

jbriggs444