Torque acting on a rotating wedge due to a linear force parallel to the axis of rotation

Let's say I have a wedge which can only be rotated about a fixed axis perpendicular to the ground. There is a cam follower mechanism constrained to vertical movement with a ball roller on the end which is applying a vertical force on this wedge at a fixed distance L from the center of rotation. As a function of F, theta, and L, how do I find the resultant torque? This is not schoolwork, it's based on a nutating mechanism I'm designing for my job.

• I'm not seeing how torque is being produced in this situation if this is just a straight wedge. Is this a screw shape? – jko Jul 27 '20 at 15:45
• Sorry, I should have been clearer. Added a 3D view to illustrate it better. I think that unless you're loading it exactly aligned in the z axis (vertical) with the angle of the slope, the follower should force the wedge to rotate as it moves downward. The sketch doesn't really show z-axis rotation. In fact, the z-axis rotation angle may have to be another function input. Let's call it phi. – Locknutbushing Jul 27 '20 at 15:55
• Now it makes sense. kamran's answer below should be enough for you. But do you need to have this rotate back as the cam is raised? Otherwise there is no force causing it to return to it's previous position. You also have an issue when the cam is in contact with the highest point of contact, if everything is perfectly aligned it will just jam into the wedge. – jko Jul 27 '20 at 16:52
• No, it's not really a textbook cam mechanism. I just explained it that way to make the problem easier to understand. It's meant to lock in either theta=0 or theta=180, so jamming is actually preferred. I'm analyzing forces acting on this part and how they transfer into torque, which could lead to oscillation due to backlash in the lock. – Locknutbushing Jul 27 '20 at 23:16

Lets callthe cam force as per your sketch P. P will have 2 components acting on the surface of the wedge, $$P* cos(\theta)_ \text{paralel to surface} , \ P sin(\theta)_ \text{perpendicular to surface}$$

Bur both of these components pass through the center of rotation, meaning they will not produce any torque, hence any rotation.

If P is too big, the shaft at the center can bend or break but wont rotate.

EDIT

After you modified your question. And if we disregard the original sketch.

there is a varying lateral force that varies as the cam moves around.

This force F multiplied by L is your torque.

, $$F= Psin(theta)*sin(\alpha) \ ,$$ Sin of the local slope angle.

That angle Alph is a simple sine function, I let you figure it.

• Sorry--I just added a comment to illustrate the problem better. Please check the comments. I realize now that a 2D sketch doesn't properly describe the problem. – Locknutbushing Jul 27 '20 at 16:01
• Thanks. Seems simple enough. When phi=0, alpha=theta. When phi=180, alpha=-theta. – Locknutbushing Jul 27 '20 at 18:02
• @Lucknutbushing, i like you sketching. – kamran Jul 27 '20 at 22:11