
#1
Sep1107, 10:16 PM

P: 36

1. The problem statement, all variables and given/known data
A pulley is hung from the ceiling by a rope. A block of mass M is suspended by another rope that passes over the pulley and is attached to the wall (the figure below ). The rope fastened to the wall makes a right angle with the wall. Neglect the masses of the rope and the pulley. Find (a) the tension in the rope from which the pulley hangs and (b) the angle that the rope makes with the ceiling. 2. Relevant equations Weight = Mg a^2 + b^2 = c^2 SOH CAH TOA (for finding the angle) 3. The attempt at a solution For the angle, I'm going to guess 45 degrees, just because it seems like it would be correct (it would be a "natural" angle for a system like this to be at.) As for the tension, i know that the tension in the *rope attached to the box* is the same all the way through (=Mg) and that the tension in the *rope attached to the pulley* must be related to this somehow, probably using SOH CAH TOA to find the angle or the length of the sides of the triangle that is formed... It's a multiple choice question, the choices being: 



#2
Sep1107, 10:41 PM

P: 36

never mind, i figured it out myself...
45 degrees was correct, and i just used TAN 45 to find the lengths of the other sides of the isosceles triangle 



#3
Sep1107, 10:56 PM

P: 336





#4
Sep1207, 02:01 AM

P: 105

Box hung over pulley at an angle??
There are things you can assume, such as:
The force on the rope at the wall is equal to the tension. You can assume that vertical component of the force on the pulley is equal to the force the weight applies on the rope. I'd look at the pulley as a point with 3 forces acting on it, and solve for the angles and forces that way. 



#5
Sep1207, 03:43 AM

P: 105

hmm i rather think the equatioon as toa cah soh
haha. yea blackwyvern is rigth. that is the way to solve the Q (visit my blog!) 


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