Vectors: x/y components, magnitude, and angle direction

[matt72lsu]

Homework Statement

Three horizontal ropes are attached to a boulder and produce the pulls shown in the figure.
1) Find x and y components of pulls A,B,and C
2) Find the components of the resultant of the 3 pulls
3) Find the magnitude of the resultant pull
4) Find direction of the resultant pull

Homework Equations

The Attempt at a Solution

I was able to find x/y components of pull A, but I'm not sure how to do it for B and C. I tried using 80cos30 and 80sin30 for pull B but it did not work. Can somebody tell me what I may be doing wrong? And could someone explain how to approach part 2? Thanks!

 

[kuruman]

I tried using 80cos30 and 80sin30 for pull B but it did not work. Can somebody tell me what I may be doing wrong? And could someone explain how to approach part 2? Thanks!

Can you be a bit more specific about what component you associated with 80cos30^o and 80sin30^o? Then I can tell you what you did wrong.

As for part 2, once you do part 1 correctly, all you have to do is add all the x-components together to get the x-component of the resultant and add all the y-components together to get the y-component of the resultant.

 

[matt72lsu]

for x component i used cos and for y component i used sin. when i put in my answers it said to check my signs. so did i do it correctly and did not put neg. x component and pos y component (since x lies on the negative x axis)? and would i do the same for pull C? I hope that is more specific for you. thanks

 

[kuruman]

Perhaps the easiest way to explain this is to note that when you write the x-component of vector V as

V_x = V cosθ

angle θ is always measured with respect to the positive x axis. This will automatically take care of the positive and negative signs. In this particular case, since vector B points "up" and "to the left", it must have a positive y-component ("up") and a negative x-component ("to the left"). Vector C points "down" and "to the left" which means that both components are negative.

So, first you need to find the angles that the vectors with respect to the positive x-axis, then find their components as I have indicated. Can you do that?

 

[matt72lsu]

Oh so you wouldn't use 30 degrees for pull B? How could I find that angle in this case?

 

[kuruman]

Start on the positive x-axis and move in a circle counterclockwise until you get to vector B. By how many degrees did you move? Hint: It is 90 degrees to the y-axis, so ...

 

[matt72lsu]

120?

 

[kuruman]

Correct. Now you can calculate the components of B. What about vector C?

 

[matt72lsu]

Thanks! I will try it after class and see how that works for me.

 

[matt72lsu]

so i just want to check and make sure i understand: for pull b i'd just use sin/cos (120) x 80N? so for pull c would it be 143 degrees? do u add 53 to 90 since it is 90 degrees from the y axis? once i find all three pull components, i'd just add all the x components and y components to each other to find the resultant components, correct? Thanks so much

 

[matt72lsu]

Ok so I got pulls A and C but B is still giving me trouble. Any help would be appreciated.

 

[kuruman]

Start at the positive x-axis and go counterclockwise all the way to vector C. Since you have to cross the negative x-axis, the angle should be greater than 180^o.

 

[A7X_\m/]

Pull B: 90 degrees + 30 degrees=120 degrees
X- component: 80cos(120)=-40.0 N
Y- component: 80sin(120)=69.28 N

Pull C: 180 degrees + 53 degrees= 233 degrees
X- component: 40cos(233)= -24.07 N
Y- component: 40sin(233)= -31.95 N

Hope this helps... ^_^