Determine force that will cause cylinder to roll out of cart

AI Thread Summary
To determine the force P that causes a homogeneous cylinder to roll out of a cart, a free body diagram (FBD) of the cylinder is essential. The forces acting on the cylinder include its weight and the horizontal force from the edge of the slot, which provides a counter-clockwise torque. The correct value of P is derived from balancing the torques, leading to the conclusion that P = sqrt(3)(M+m)(g). It is advised to avoid considering reaction forces to simplify the analysis. A clear understanding of the forces and torques at play is crucial for solving the problem effectively.
bkw2694
Messages
33
Reaction score
0

Homework Statement


I've uploaded an image of the problem below. The problem states: Determine the value of P which will cause the homogeneous cylinder to begin to roll up out of its rectangular recess.The mass of the cylinder is m, the mass of the cart is M. Friction of the cart wheels i negligible.

Homework Equations



ΣF_x = ma_x
ΣF_y = ma_y
ΣM_G = 0

The Attempt at a Solution



I'm really not sure how to approach the cylinder in this problem, but I can see that a FBD of the problem shows m*g and M*g going downward in the y-direction, and P is in the x-direction. I'm also assuming the cart is accelerating in the x-direction, which means a_y = 0.

The correct answer is P = sqrt(3)(M+m)(g). I'm obviously missing some key force because I keep getting P =0.
 

Attachments

  • 6.8.JPG
    6.8.JPG
    14.6 KB · Views: 901
Physics news on Phys.org
Have you drawn a free body diagram for the cylinder alone? What forces act on the cylinder when it is just at the verge of tipping up and out?
 
The edge of the slot provides the horizontal force that accelerates the cylinder.
Generally, its best to avoid reaction forces - in this case it seems simpler to do so.
The reaction force on the C. M. provides a counter-clockwise torque about the edge of the slot.
The weight of the cylinder provides a clockwise torque about the same edge.
Obviously, once the cylinder starts to tip (torques are equal) the counterclockwise torque will exceed the clockwise torque.
 
J Hann said:
The edge of the slot provides the horizontal force that accelerates the cylinder.
Generally, its best to avoid reaction forces - in this case it seems simpler to do so.
The reaction force on the C. M. provides a counter-clockwise torque about the edge of the slot.
The weight of the cylinder provides a clockwise torque about the same edge.
Obviously, once the cylinder starts to tip (torques are equal) the counterclockwise torque will exceed the clockwise torque.
The approach that I have in mind is different from this. It ends with the same result but avoids talking about torque, angular momentum or reaction forces. But either way, bkw2694 needs to do that free body diagram for the cylinder alone.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top