Recent content by Valerie Prowse

Ah, I see the problem. This question is from a previous exam, but my course did not cover this topic, hence why I had no idea what you meant by power. I went through my textbook and found the section on power, which was not assigned, but I will give it a quick look over and try this question...

I'm not sure what you mean by that..

Homework Statement A loaded truck has a mass of 3100 kg. The maximum speed it can maintain on a 5.0° hill is 80 km/h. What constant speed could the truck maintain on a hill with a slope of 10.0°? Assume the total force due to air resistance and friction is 700 N and that it does not vary with...

Great! Thank you everyone for your help! :woot:

I worked out the rest of the question in case this was okay: ∑Fy = FNy - mg - FT⋅sin28 FNy = FT⋅sin28 + mg = (984.8⋅sin28) + (40⋅9.8) = 462.3 + 392 = 854.3N ∑Fx = FT⋅cos28 + FNx - Fp = 0 Fp - FT⋅cos28 = FNx FNx = 520 - (984.8⋅cos28) = -349.5 (This makes me think that FNx is pointing left...

Ahh this makes sense. So, the torque equation about the bottom of the beam would be: Fp = 520N ∑τ = 0 = Fp ⋅ 5m - FT⋅cos28⋅3m - FN(x)⋅0m 0 = Fp ⋅ 5m - FT⋅cos28⋅3m FT = Fp⋅5 / cos28⋅3 FT = 2600/2.64 = 984.8N However, this is much more than Fp... and the horizontal component of FT would also be...

Does this mean that the reaction force is different from the normal force? I did draw the FBD, but I am unsure in which direction the reaction force would be pointing. I thought maybe, from looking at different types of questions, that it would be pointing in the direction of the cable? Or...

Homework Statement A uniform vertical beam of mass 40 kg is acted on by a horizontal force of 520 N at its top and is held, in the vertical position, by a cable as shown. a. Draw a free-body diagram for the beam, clearly labeling all of the forces acting on it. b. Calculate the tension in the...

So you mean torque? If τ = rF⊥, τ = (L/2)(mg) = 58.8 and α = τ/I = τ/(1/3mL2) = 14.7 rad/s

Whoops, my bad. I misread part a. Instead... ω2 = ω02 + aα(θ-θ0) α = (ω2)/2θ α = (5.4)2/ 2(1.57) = 9.28 rad/s2 I got the value 1.57 rad = 90° from the attached image.

Sorry, I didn't include subscripts. The angular velocity on the left in that equation is the angular velocity at point B, which I found in part A to be 5.4 rad/s, assuming that is the correct answer. The angular velocity on the right is unknown.

Homework Statement Consider a uniform rod of mass 12 kg and length l.0 m. At its end, the rod is attached to a fixed, friction-free pivot. Initially the rod is balanced vertically above the pivot and begins to fall (from rest) as shown in the diagram. Determine, a. the angular acceleration of...

Thank you everyone for your help! :smile::smile:

How is this? If α = a/r, I can use that to plug into α in the equation? if so, I THINK I have it: 2rm(g-a) / (r^2 * α) = M 2rm(g-a) / (r^2 * [a/r]) = M which becomes 2r^2m(g-a) / (r^2 * a) = M and cancel out...

The reason I am avoiding this is because we haven't yet been introduced to rotational energy (next lesson), so I am unfamiliar with how to use the information. But I appreciate the help! :)