jonesy01
Mechanical
- Mar 17, 2008
- 12
Need help, due to safety issues I need to find out what force would be needed to tip over a cart.
I am analyzing force required to tip over a cart and the cart consists of 3 items:
-The cart with wheels = (148 lbs)
-Tank w/water = (370 lbs)
-Pump = (20 lbs)
The way I am analyzing this is if the front wheel collides with something what’s the force required for the cart to tilt or tip forward. I believe the cart will tip if the torque at the wheel exceeds the torque due to the total weight of the cart.
-1st, I found the center of gravity for the cart with all components. Multiplied the weight of each component times their midpoint distance from the front of the cart to get their moments. Added them all together.
pump, 3.87 in x 20 lb = 77.4 lb/in
tank, 20 in x 370 lb = 7400 lb/in
cart, 18.5 in x 148 lb = 2738 lb/in
----------------
10215.4 lb/in
-2nd, got the sum for the weight of the cart and all components = 538 lb and then divided the moment sum by the weight and got 18.98 in as the center of gravity from the front of the cart.
-3rd, the only way the cart will tip is if the torque at the front wheels exceeds the torque due to the weight of the cart. Formula I used:
TORQUEcw = TORQUEccw
(f)(d) = (f)(d)
(f)(41.56”) = (538 lb)(18.98”)
f = 245 lb, so 245 lbs needed to tip over the cart if the front wheels at point A hit something and if the cart with all items weighs 538 lbs.
Am I missing anything here. Any advise is much appreciated.
I am analyzing force required to tip over a cart and the cart consists of 3 items:
-The cart with wheels = (148 lbs)
-Tank w/water = (370 lbs)
-Pump = (20 lbs)
The way I am analyzing this is if the front wheel collides with something what’s the force required for the cart to tilt or tip forward. I believe the cart will tip if the torque at the wheel exceeds the torque due to the total weight of the cart.
-1st, I found the center of gravity for the cart with all components. Multiplied the weight of each component times their midpoint distance from the front of the cart to get their moments. Added them all together.
pump, 3.87 in x 20 lb = 77.4 lb/in
tank, 20 in x 370 lb = 7400 lb/in
cart, 18.5 in x 148 lb = 2738 lb/in
----------------
10215.4 lb/in
-2nd, got the sum for the weight of the cart and all components = 538 lb and then divided the moment sum by the weight and got 18.98 in as the center of gravity from the front of the cart.
-3rd, the only way the cart will tip is if the torque at the front wheels exceeds the torque due to the weight of the cart. Formula I used:
TORQUEcw = TORQUEccw
(f)(d) = (f)(d)
(f)(41.56”) = (538 lb)(18.98”)
f = 245 lb, so 245 lbs needed to tip over the cart if the front wheels at point A hit something and if the cart with all items weighs 538 lbs.
Am I missing anything here. Any advise is much appreciated.