Resolving Forces to Ensure Motor Strength: A Screw's Tale

[Ahatem]

So I have a motor and this motor will turn a screw and bring a platform down to expose fairings during rocket flight. The platform is connected to the fairings. Due to drag pushing the fairings down there will be an upwards force pushing the platform against the screw holding it down.

The only way for the platform to go back up is to spin the screw in reverse.

So I have the threads of a screw being exposed to an upward force. When I resolve the forces I see that no torque will cause the screw to spin. However if there is a torque there, I want to make sure that the motor can handle it.

My approach: I resolve the upward force into 2 components. One along the thread and one perpendicular to the thread. Then resolve each of those onto the vertical and horizontal axes. Obviously enough they cancel out and no force in the horizontal axis remains. (knew this would happen since no initial force in horizontal but just wanted to double check).

So is there a torque that my motor needs to withstand? If so, where does it come from?

Someone suggested that only the component perpendicular to the thread should be considered, and the component along the thread should be ignored. That leaves a force in the horizontal of the form Fcos(x)sin(x).

 

[Simon Bridge]

Usually a screw is threaded at so shallow an angle that static friction is enough to hold the screw in place against any likely load.

In the imaginary frictionless screw-hole, a force along the axis of the screw will cause it to turn... it is a basic mass-on-a-slope problem as I think you have surmised.

 

[phinds]

What you REALLY need to watch out for is vibration. In the presence of vibration a screw can "unscrew" with very little force --- the vibration just has to be strong enought to overcome the force at the peak of each vibration cycle --- if it does that, the screw will unscrew.