There is a box with a width of 2a, a height of 2b, and a mass of m. There is a rope pulling the box from the one of the upper corners.
(a) assume the floor is frictionless. What is the maximum force that can be applied without causing the box to tip over. hint: use the center of mass
(b) Repeat part (a) but now let the coeffiecent of friction be mu.
(c) What happens to your anwer to part (b) when the box is sufficiently tall? How do you interpret this?
Torque= Force * distance from center of mass
Tension = F(rope) + Force(friction)
Statics: the total torque must equal zero.
The Attempt at a Solution
Part (a) was easy.
max torque = 0 = Torque mass + torque rope
0 = F * sin 90 +F(rope)*sin angle
using the center of mass sin ∅ = a/(b^2+a^2)^0.5
using the center of mass cos ∅ = b/(b^2+a^2)^0.5
using the center of mass tan ∅ = a/b
0=F*cos ∅-mg; rope vs block
Tension(rope) = F*sin ∅
Tension = mg/cos∅ *sin∅
"Tension = mg (a/b)"
Part (b) has me a little stumped as I am not sure about the amount of torque generated by the friction
The torque is close to the same equation but the friction is going to have the same rotation as the rope. So Total torque is = to Torque mass + torque rope+torque friction. I tried the F*Mu didn't work so, that is where I became stuck.