Recent content by h8torade

I found the torque of the applied force which is 1.12268(225sin18)=78.05 but I just do not know where to go from here.

So the torque of friction should equal the torque of the applied force in order to maintain equilibrium. So I need to find the torque of the x and y components of the applied force to find the applied force's torque and then divide that by the force of friction in order to get the distance.

Homework Statement A 2D rectangular block is being pulled at a constant speed with a height of 2.2m and width of 51cm and weighs 403N. The pulling force of 225N is being applied at an 18 degree angle at a height of .666cm from the base of the box. What distance from the lower left corner...

so the answer to the stress is 1685.6 N / .0000143 m^2 = 1.17874e8 which makes E = 1.63714e11

Homework Statement A load of 172 kg is supported by a wire of length 2.07m and cross-sectional area .143cm^2. The wire is stretched by 0.149 cm. Find tensile stress, strain, and Young's modulus. Homework Equations stress= F/A strain= ∆L/L E= stress/strain The Attempt at a...