Recent content by alco19357

Yes, I think the book didn't use the rounded answer for the tension force when they calculated the normal force. Hence, they would get an answer around like 18.5 or so

Did I do everything right to find the coefficient of kinetic friction then? wWhere did I go wrong in that solution thanks

LT*sin160 - .5*L*W*sin130 = SUM OF TORQUES = 0 LT*sin160 = .5*L*W*sin130 T = .5 * W * (sin130)/(sin160) The website said that was the right answer

Thanks for the help, I got T = 2300 Now for part b though, I would think the coefficient should be u=f/N f is R_x N is R_y R_y + Tsin60 - W = 0 R_y = W - Tsin60 ...= 210*9.8 - 2300*sin60 ...= 66.14157 R_x - Tcos60 = 0 R_x = Tcos60 ...= 2300*cos60 ...= 1150.000 u = f/N u = R_x / R_y u =...

Okay I see that. Now I got: (R is the reactionary normal force of the ground, w is weight, L is length of the beam) R_x + Tcos60 = 0 R_y + Tsin60 - w = 0 T_1 = -L/2 * W * sin130 (counter clockwise is +) T_2 = L*T*sin160 Are these two tensions right? If so, I'll have three equations 1. R_x +...

grzz, thanks for the reply I don't understand how it's not 40. Isn't there a force, T that the beam exerts back on the cable? Shouldn't those forces be opposite but equal?

Homework Statement A uniform, 210-kg beam is supported by a cable connected to the ceiling, as shown in the figure . The lower end of the beam rests on the floor. a. What is the tension in the cable? b. What is the minimum coefficient of static friction between the beam and the floor...

Nevermind, using T = (mv^2) / r, and setting L1 = L2, I found v to be (v1*r1) / r Plugged that into the tension and found T = (m*v1^2*r1^2)/r^3

Homework Statement A block with mass m is revolving with linear speed v1 in a circle of radius r1 on a frictionless horizontal surface (see the figure ). The string is slowly pulled from below until the radius of the circle in which the block is revolving is reduced to r2. Calculate the...

Nevermind. For those of you who are having trouble, figure I as a numerical number. So I = m*R^2 + 6/3 * m*R^2 Note that the R is the radius of the wheel, hence, R is the same for both parts of the equation. However, m in the first part is the mass of the WHEEL (so you'll take the...

Homework Statement A 45.0-cm diameter wheel, consisting of a rim and six spokes, is constructed from a thin rigid plastic material having a linear mass density of 25.0 g/cm. This wheel is released from rest at the top of a hill 52.0m high. a.) How fast is it rolling when it reaches the...

Thank you so much! :smile: That's the right answer!

I think what you're saying is that I have: [(m^3) / (s^2)] - [(m^2) / (s^2)] So am I missing the d variable? So should it be: (m_B * g * d) - (u_k * m_A * g * d)

Thank you for the help. I applied what you said and got the following answer: What have I done wrong? Thanks

1. I don't know. I thought maybe that the velocity of a is non existent, but wouldn't the velocity of a equal the velocity of b? 2. I would say since the frictional work done is between the surface and object a, that mass should be the mass of block a. Thank you for your help