How Much Force Is Needed to Tip a Loaded Cart?

[cmcpolo]

Attached is a sketch of a sheet rock cart. The cart weighs 79 pounds and I have indicated the location of the centroid. The load on the cart consists of several sheets of material that weigh 1200 pounds. I am trying to determine the pulling force required (F) to tip the cart over (assuming the hinge point A doesn't slide when the cart is pulled).

A little background...I am a professional engineer and I am trying to help a client but it has been quite a while since I took physics or statics. I have arrived at an answer of 87 lbs but was hoping for a check from someone with greater knowledge.

Thanks for any help.

 

[Redbelly98]

Welcome to Physics Forums. I get 134 pounds.

For me, it's easier to think in terms of how much torque is required to tip the cart, and then calculate the required force given its 3.07 ft height above the pivot point A.

Your drawing indicates that the cart centroid is 0.46 ft to the right of point A, so the cart contributes 0.46*79 = 36 ft-lbs to the required torque.

The horizontal distance from point A to the sheet rock's centroid is

0.66' - (2')*cos(80°) = 0.31 ft​

so the sheetrock requires a torque of 0.31*1200 = 375 ft-lbs.

The total toque required is 36+375 = 411 ft-lbs.

Given the 3.07' height of the force, a force of (411/3.07)=134 lbs is required to tip the cart.

Please make sure you follow my calculation, in particular where I calculate the location of the sheetrock's centroid.

Final note: the calculation is quite sensitive to the accuracy of the 80° figure. If it were actually 75°, the force becomes 67 lbs; for 85° it's 200 lbs.

Hope this helps.

 

[cmcpolo]

Redbelly 98-
Thanks for taking that time to run through that calculation. After a bit more research after my post, i realized that I had incorrectly shown other forces in my sketch as they were internal forces. The correct calculation, as you have indicated, is the moment calculation with the centroids.

Thanks