Tension. Application of Newton's Laws.

[valeriex0x]

Homework Statement

Two weights are hanging by the following cables, A, B, C.
1. Find the tension in Cable A.
2. Find the tension in Cables B and C.

-----ceiling----------
\ 60) (60 /
\ /
\ /
B \ / C
\ /
\ /
[_] small box:72 N
|
A |
~~~
| | Big Box: 150 N
~~~

Homework Equations

Free body diagram of big box, maybe some trig for finding cables b or c

The Attempt at a Solution

1.
I know that the big box has a Force:weight pointing down of 150 N, and normally, if it was resting on a flat surface it would have normal force pointing up 150N. But because it is suspended by cable A from small box 72N, I'm not sure how to approach finding the tension in cable A. Would it be okay to think the tension in cable A is 150 N + 72 N?

2.
I am thinking I could draw a resultant? and that its magnitude would be 72 N, given by the box hanging from it. and from there do Opp/hyp or Sin (60).
-----ceiling----------
\ 60) | (60 /
\ | /
\ | /
B \ | / C
\ | /
\ |/
[_]



Any help would be appreciated! thanks

 

[valeriex0x]

The figure is supposed to be an a triangle with its base parallel with the ceiling and its other two sides pointing into a V .

 

[Ignea_unda]

For cable A try thinking of the small box as a "ceiling" with only cable A below it and from that cable, the large box would be hanging. What would you say the tension is in that situation? Is this situation any different?

For part (b) don't forget that you need to include both the small box's weight (force) AND the large box's weight (force).

 

[valeriex0x]

1. So to find the tension in Cable A, I take the 72 N pointing down, and the 150 N pointing up, and get a net force of 78 N. Would that be correct. I feel confused into thinking that the sum of forces should be zero if these boxes are hanging in equilibrium.?

2. 72 N + 150 N= 222N(sin(60))= 192 N the tension in cable B!>??

 

[Ignea_unda]

1. So to find the tension in Cable A, I take the 72 N pointing down, and the 150 N pointing up, and get a net force of 78 N. Would that be correct. I feel confused into thinking that the sum of forces should be zero if these boxes are hanging in equilibrium.?

2. 72 N + 150 N= 222N(sin(60))= 192 N the tension in cable B!>??

I guess I wasn't clear enough. For part (a), nothing is moving. Because nothing is moving, the only "force" being applied on the lower string is that of the large box. It is the same as if the large box were hanging from the ceiling from cable A and you were asked what the tension is in that cable.

For part 2, don't forget that you have 2 cables. Would cable B be taking all the load? Think of what you can do to take advantage of the geometry in the problem.

 

[valeriex0x]

1. The tension in Cable A is 150 N.
2. The tension in Cable B=Cable C = 128 N.

2) 72 N + 150 N= 222 N
Sin(60)= opp/hyp=222N/x=.8660254
Put sin(60) over one, cross multiply, solve for x (Cable B) x= 256 N

but that is the total tension.

Total Tension 256N divide by 2 = 128N tension in cable b and c

??! I think i got it!