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# At what least height above the floor must a horizontal force of magnitude 390 N be applied to the crate?

A uniform cubical crate is 0.700 m on each side and weighs 570 N. It rests on the floor with one edge against a very small, fixed obstruction. At what least height above the floor must a horizontal force of magnitude 390 N be applied to the crate to tip it?

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- SimonLv 78 years agoFavorite Answer
This is a torque problem. Just set the torque needed to move the center of mass equal to the applied torque.

- Anonymous4 years ago
unknown = a Forces in this question : Fg (given) , Fn, Ffr, Fp Fg = Fn (element floor) = 1020N Ffr=kFn (ok being ur coefficient of friction, i dont have mew) Fnet = Fp-Ffr = ma Fp= (390N)(cos 29)= 341.a million N m=Fg/g = 1020N/9.8= 104.08kg Ffr= (0.25)(1020N)= 225N 341.1N- 225N = (104.08kg)a (< this equation from Fnet) remedy for a : a = 0.83m/s^2

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