- #1
Hans Herland
- 17
- 1
Homework Statement
As the title states, the goal of this problem is to determine the weight of an object based on the torque caused by a force applied to it. The object in question is a wheelbarrow, and the force applied is a person lifting it by the handle using 100N with an angle of -30° from the x-axis. Hopefully the image I'm providing gives a sense of the problem.
The resultant torque caused by the force in point B combined with the torque caused by the weight of point G to point A is 0.
Sorry for completely ruining the scaling of distances, but the numbers are correct. =)
Homework Equations
For the torque equations so far we've been using: ΣM = Fxdy - Fydx
For G, the equation is: G = mg. That is, at least the one I think would be relevant here.
The Attempt at a Solution
What I've attempted so far is to first calculate the torque caused by the applied force in point B. I am a little uncertain as to how the length of the arm works in that equation. Is, for instance, the length of arm Bx -1.2, or do you just use 1.2 without taking into consideration what direction the arm is?
(100 cos 30)(-0.65) - (100 sin 30)(-1.2) = 3.7Nm
So far, I feel like I'm on the correct path. With this answer, I start looking at the torque in point A, and that's where I start feeling unsure.
(FAx)(0.5) - (FAy)(0.3) = -3.7Nm
Here I have two unknowns, since I don't know the actual force on point A. What I've tried to do here is to replace those unknowns with the mass of point G, mg, but do I need to apply that with trigonometry?
The angle from point A to G is 59° when I use tan-1 (0.5/0.3).
(mg cos 59)(0.5) - (mg sin 59)(0.3) = -3.7
m(9.81 cos 59(0.5)) - m(9.81 sin 59(0.3)) = -3.7
m(2.526 - 2.522) = -3.7
m = -3.7 / (2.526 - 2.522) = 925kg
I've done some other attempts, but I don't get as far as I feel I do here. 925kg doesn't sound like a reasonable answer, so there's something I'm not getting here.
I'd be grateful for any suggestions. =)
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