Recent content by RioAlvarado

But my mass is already based on there being a distance of one meter from B to the block. I say since it is a uniform bar we know that in the one meter of bar there is 260/3 kilograms present. Isn't it 1.5 meters? which is what I had in my initial equation, I don't recall ever putting in 2.0.

Alright, trying to work more carefully now: ## 1∗(\frac{260}{3}∗9.8)+1∗(60∗9.8) = 0*F_B + 1.5*F_A##. The distance from B is zero, so if I use that as the reference point its torque is zero, it is still one meter to the point where the red block is, so the ones stay there. And point A is 1.5...

OK. So if we chose the particular point as B our torque equation would be ##1 * (\frac{260}{3} * 9.8) + 1*(60*9.8) = 1.5*F_B + .5 * F_A ##. Am I on the right track?

Homework Statement A horizontal uniform bar of mass 260 kg and length L = 3.0 m is placed on two supports, labeled A and B, located as shown in the diagram. A block of mass 60 kg is placed on the right end of the bar. Find the normal forces FA and FB exerted on the bar by the supports. Enter...

Sorry I couldn't make my solution more clear. I'm not surprised it isn't the best solution, however it did give me precise, and correct answers, and I'm happy with that. If you really think I should elaborate more on my answers for the benefit of those who might be working this problem I could...

Alright, for a. I realized that I needed to consider the alternate angle to the 40, that is 50 degrees. I then said ##tan(50)= \frac{900*25}{700*v_2}## and solved. For speed of the car afterwards I used ##(700+900)*v = \frac{700*v_2}{cos(50)}## and solved that, for the impulse I used the model...

I thank you for your reply, but as it turns out, while I was awaiting responses I continued working on the problem and managed to figure it out. Nevertheless I thank you very much for taking the time to reply and offer assistance.

Homework Statement [/B] Two cars are approaching a perpendicular intersection without a stop sign. Car 1 has a mass ## m_1 = 900 kg ## and is heading north at ##v_1 = 25 m/s ##. Car 2 has mass ##m_2 = 700 kg ## and is heading west at an unknown speed ##v_2##. The two cars collide at the...

My eternal thanks good sir or madame. Your advice was instrumental in my finding the solution.

Should have been ##=1.2ma## I think.

So it would then be: ##2f - 1.2mg - 2R = 1.2mg## (because since before you defined the force R to be the force the platform exerts on the men upwards, the normal force for the platform by the men should be twice the negative of R).

Alright, I think I understand what you mean.So would the equation for the other man just be ##R+f−mg=ma## since he is identical to the first man and the equation for platform would be ##2f -1.2mg = ma## Two f for the tensions forces acting on the block, minus the downward component of force...

Wouldn't the force R simply be equal the normal force the man is exerting on the platform and in turn then be equal to ##mg##, giving ##f = ma## out again since R would cancel with ##mg##.

Sorry, that was a typo which persisted through my idiocy. I corrected it.

Homework Statement Two construction workers each of mass m raise themselves on a hanging platform using pulleys as shown above. If the platform has a mass of 1.2⋅m, the initial distance between the pulleys and the platform is d, and the workers each pull with a force f on the ropes, what is...