It is getting clearer now. I have one more question. Why do we calculate friction on the wheels without brakes? I thought that the braked ones are responsible for slipping resistance.
The centrifugal force is equal and opposite to the centripetal one. Therefore, including the centripetal force into the reaction (which is wrong) would arithmetically give the same result as calculating the reaction based on FT and FR and then adding the centrifugal force. So, it is "just" a...
Yes, the front is on the right. So, if the rolling resistance gets outweighed by the reaction on the trolley, the device slides even though a friction force on the braked wheels is greater then the reaction? It's counterintuitive to me.
Yes, that's right.
Then, if I understand it correctly, Fc...
Thanks a lot! Your question has made me aware of some issues. I've decided to change the conditions. Now, the rate of rotation varies, but instead the motor's torque (T) is constant. An updated force diagram looks like this:
***EDIT***
FT should be pointed in the opposite direction - my bad...
Let's assume that in the beginning the platform moves with an angular acceleration ε to achieve the rotation rate ω. Therefore, the CoG moves with a tangential acceleration a = ε*R. Horizontal component: ax = a * cosα. Vertical component: ax = a * sinα.
The friction force decreases, because...
Yes, I was thinking of the static coefficient. Sorry I didn't put it clearly.
Rotation is achieved by means of a torque applied to the platform's axis of rotation. The torque is generated through a worm gearbox, there is no force acting directly on the platform. Should I decompose the torque to...
Homework Statement
I'm designing a device for changing a load position from vertical to horizontal. It has a wheeled frame (2) which allows an operator to transport the load after reorientation.
The combined centre of gravity of a rotating platform (1) and the load moves forward due to...