Finding Horizontal Force in a Crank at 10N

[richard9678]

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.

 

[Nugatory]

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]

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.

This is mistaken. A force of 10N on a diagonal does not break into halves at 5N horizontal and 5N vertical.

 

[richard9678]

OK. So, cosθ or sinθ is the correct trig. ratio to use to establish force provided by a crank lever, in the vertical or horizontal as the case may be. Never fully appreciated that at 45 degrees the tangential force does not simply divide into a half of the force at a tangent to the crank. I do now. Thanks.