Calculate Cable Tension in a Slanted and Horizontal Position

[bamb3ry]

Tension help?

Consider the 59 N weight held by two cables. The left-hand cable is horizontal.

What is the tension in the cable slanted
at an angle of 51◦? Answer in units of N.

What is the tension in the horizontal cable?
Answer in units of N

 

[Dick]

You have three force vectors pointed out from the point where the cables meet. The sum must be zero. Now split the forces into x-y components and equate the horizontal and vertical forces. Now stop double posting. Now start showing an attempt at solving the problem if you expect significant help.

 

[bamb3ry]

Sorry for double posting but, I was absent today in school when we learned this, and my teacher expects me to know how to do this. My grandmother died and i had to go to her funeral. What do you mean by splitting them into x-y components, and isn't their only 2 vectors pointed out

 

[Dick]

There are three forces. One to the left that's horizontal. One the the right that's upwards at a 51 degree angle. And one that's downwards with a magnitude of 59N. The vertical forces have to cancel first. Only two of them have a vertical component. Please don't say you don't know how to split a force into vertical and horizontal components. Use sin(51 degrees) and cos(51 degrees), ok? Sorry about your grandmother.

 

[bamb3ry]

Nah it's fine and thanks for helping me though.51sin= .777145 51cos= .6293

Now after i split them what do i have to do with the numbers?

 

[Dick]

Call the magnitude of the 51 degree force F. Then the upward part of F is F*sin(51 degrees), assuming 51 degrees is the angle from the horizonal, isn't it? So isn't F*sin(51 degrees)=59N? Since the vertical components have to balance? Solve for F. Now balance the horizontal components. Sorry, I've got to zzzzz now.

 

[tiny-tim]

welcome to pf!

hi bamb3ry! welcome to pf!

Nah it's fine and thanks for helping me though.


51sin= .777145 51cos= .6293

Now after i split them what do i have to do with the numbers?

i'll say this quietly so as not to wake Dick :zzz: …

in the class you missed, the teacher probably said that forces are vectors, and so when you add forces, you must add them like vectors ("vector addition") …

you can add vectors either by using a vector triangle

or by taking the x and y components, and adding them separately …

the three forces (two unknown tensions and the known weight) must add to zero, so the x and y components will give you two equations wiht two unknowns, which you should be able to solve …

shhh … ​

​

 

[Dick]

Thanks for keeping the volume down! I didn't hear a thing. Yaawwwn.

 

[tiny-tim]

Thanks for keeping the volume down! I didn't hear a thing. Yaawwwn.

oooh … can we make a NOISE now?

… so long as we don't fight!​