Recent content by rml4589

WOOO! Thank you SO much lol. I've seriously been working on that problem for 3 days now. it was the last of my pre-final practice test. It was only worth like 1/1000 of my grade, but it will definitely help me on the final tomorrow. We got snowed in here in Mass so they moved the final to...

so Rtension = length of the BEAM not the cable? and it doesn't cancel out because Rweight = 2Rtension. SO 137.2N * 2m * .656 deg. = Ftension * 4m * .755 and Ftension = 59.6 N? please tell me I'm finally right haha

as a pre-ps, thanks for putting up with me as long as you have haha Rweight = 2 because the weight acts on the middle of the beam. as for Rtension, like you said, acts on the end of the beam. so wouldn't it be the distance from the wall to the end of the beam, which happens to be Ltension...

ok, so using your suggestion, the formula is: Fweight*Rweight* sin41 = Ftension*Rweight * sin49? are the Rweight supposed to cancel? if so, Fweight * sin41 = Ftension*sin49 137.2 * sin41 = Ftension*sin49 Ftension = 119.3 N?

ok you lost me :( so Rweight is equal for both sides of the equation as well as Fweight? or is Rtension = Rbeam? i'm confusing myself i think

4.6 was the length of the cable not the beam. since i know the angle between the wall and the beam, and i knew the length of the beam (the hypotenuse), i used sin41 = Lcable / 4... which i just calculated to be 2.62 haha wow strange mistake there... ok so 137.2N * 2m * .656 = Fcable * 2.62 *...

ok so it would be 137.2N * 2m * .656 deg = Fcable * 4.6 * .755 Fcable = (137.2N*2m*.656) / 4.6 * .755 Fcable = 51.83N ?

Rbeam = 2m Rcable = 3.02/sin41 = 4.6 / 2 = 2.3 so it would be, with everything plugged in Fbeam*Rbeam*sin(41) = -Fcable*Rcable*sin(49) and therefore 137.2N * 2m * .656 deg. = -Fcable * 2.3m * sin(49) solving for Fcable Fcable = -(137.2N * 2m * .656 deg.) / (2.3m * .755 deg.)...

ok i think i know what you are saying. the torque of the cable will be equal but opposite the torque of the beam, causing a net torque of 0. using the torque of the cable i should be able to figure out the tension of the cable. since tension is just a force, and force is just ma, in this case...

the only torque about that point would be the torque of the beam right? so the torque of the beam would be offset by the tension in the cable? sooo if i fiind the torque of the beam, that will equal the tension in the cable or am i making up my own laws of physics here?

i was trying to figure it out using the net forces = 0. if i chose the pivot point being the point where the beam touches the wall, would i have to figure it out in the x and y directions? or could i just take the d to be the length of the beam?

Homework Statement The beam is 4.0 m long and has a mass of 14.0 kg. If the beam makes an angle of 41.0 degrees with the wall, what is the tension in the cable? The attachment is the picture to go along with visualizing the problem. Homework Equations sum forces = 0 sum torqes = 0 The...