Find the tension in the cable BE and CF and the reaction at D.
if a 900N load was placed at A what would be the tension in the cables?
If both cables were made from the same material and had the same thickness. which cable would break first?

Determine the reaction forces exerting on the pulley at point B (there is no friction, B is a pulley)

- Thanks in advance for the help

2. Originally Posted by iamdenis

Find the tension in the cable BE and CF and the reaction at D.
if a 900N load was placed at A what would be the tension in the cables?
If both cables were made from the same material and had the same thickness. which cable would break first?
I assume you know mechanics, free-body diagram (FBD), moments, etc.

The weight on the drawing is 600N. In the text it is 900N. Which one?
Let us say it is 900N.

First get the vertical, Fy, and horizontal , Fx, components of the two tensions.
Say T is the tension at BE, and U is the tension at CF.

For the T:
T is acting along BE, so get the length of BE.
BE = sqrt[(200^2 +80^2) = 215.4066 mm
By proportion,
Tx = T(200/215.4066) = (0.92848)T
Ty = T(80/215.4066) = (0.37139)T

For the U:
CF = sqrt(100^2 +80^2) = 128.0625 mm
So,
Ux = U(100/128.0625) = (0.78087)U
Uy = U(80/128.0625) = (0.62469)U

Now, to solve for T, cut the system vertically between points C and D to create a FBD. Take moments at point C. Break the T into its Tx and Ty components.
The Tx has no moment at point C because its line of direction passes through point C.
Ty(100) = 900(200)
TY = 1800 N
So, T = 1800 / 0.37139 = 4847 N tension --------------answer.

For the U,
in the same FBD above, break the U into its Ux and Uy components.
Take moments at point B. Ux has no moment there
Uy(100) = 900(100)
Uy = 900 N
So,
U = 900 / 0.62469 = 1441 N tension ----------answer.

For the reaction at point D:
Since the beam is supported by a roller there, there will be no vertical restraint there. It will all be horizontal.
It will be Ux.
It will be 1441(0.78087) = 1125 N compression -----answer. (Oppss, my mistake here. Let me correct it.)
I forgot the the Tx.
The reaction at D should be R = Tx +Ux
R = 4847(0.92848) +1125 = 5625 N compression ---------corrected answer.

Which cable will break first?
Cable BE, because it carries more stress than cable CF. ------answer.
Or, because the T is greater than the U.

3. you're awesome, thanks. about the 600, you basically have to solve 600 and then 900. but thats easy. thanks again

4. Originally Posted by iamdenis

Determine the reaction forces exerting on the pulley at point B (there is no friction, B is a pulley)

- Thanks in advance for the help
Create a FBD for the lower pulley, pulley A, by cutting the two cords from pulley B.
Let us say we call the tension at the cord to the left of A as P,
and the tension in each of the cords to the right of A as U....they are equal.

For equilibrium, summation of vertical forces at A is zero, so,
P*sin(alpha) +2[U*cos(beta)] = (160 kg)(9.8) = 1568 N ----------(1)

Summation of horizontal forces at A is zero, so,
P*cos(alpha) = 2[u*sin(beta)] --------------(2)

Solve those two equqtions simultaneously to get the Q.
Then, the reaction at the pulley B is the pull of the two U's.
If you want the vertical and horizontal components of the reaction at B, it should be easy for you.