Solving A Rocket & Traction Apparatus Challenge

[sloan13]

Homework Statement

A) A rocket fires two engines simultaneously. One produces a thrust of 675N directly forward while the other gives a thrust of 450N at an angle 20.4∘ above the forward direction.

a) Find the magnitude of the resultant force which these engines exert on the rocket.
b) Find the deviation of the direction (relative to the forward direction) of the resultant force which these engines exert on the rocket.

B) A patient with a dislocated shoulder is put into a traction apparatus as shown in the figure. The pulls A⃗ and B⃗ have equal magnitudes and must combine to produce an outward traction force of 5.80N on the patient's arm.

There is a picture of the arm out. There is a horizontal line then two vectors 32 degrees above and below the line.

2. The attempt at a solution
A)

a) 954 N
b) 16.8 degrees

B)

a) ∣∣A⃗ ∣∣3.3 N

I am stumped. Thank you for any input!

 

[jackarms]

If you could show your work for part A, that'd be great to see if your reasoning is right. And then I think I understand what the situation looks like for B, but I'm not quite sure what the question is asking, so if you could clarify that as well.

 

[sloan13]

Honestly it was guess work. There aren't any examples like this in our book and we skipped it in class. (this is online homework)

 

[jackarms]

Well, the first part involves finding a resultant force, so you have to find the components of the two vectors given using:

$v_{x} = |v| \times cos(\theta)$ and $v_{y} = |v| \times sin(\theta)$

And then add them together, and then find the magnitude using:

$|v| = \sqrt{v_{x}^{2} + v_{y}^{2}}$

Does all that look familiar?

 

[BvU]

Suppose the 450 N was 675 N. The angle would be 10.2 degrees then, wouldn't it ? If the force isn't 675 N but a mere 450 N, the resultant angle can't be bigger than 10.2 degrees, can it? IMHO you should guess again...

Make a drawing. Show it. And like Jack I wonder what the question is for part B. My guess for the answer would be a a few % more than 3.3 N. And 5.8 N isn't enough to fix a dislocated shoulder by far !

 

[sloan13]

To jack: yes I know those formulas but I thought you were supposed to do something else. Do I just add them as usual?

 

[jackarms]

To jack: yes I know those formulas but I thought you were supposed to do something else. Do I just add them as usual?

Yes, it's just vector addition as usual. The key part of the problem is that it gives you vectors in terms of forces. The thrust of 675N is one vector, with magnitude of 675N at 0° above the forward direction, and the other is a vector with magnitude 450N at 20.4° above the forward direction. Since both engines are fired at the same time, the two forces can be added into a net force with vector addition.

 

[sloan13]

Ok. I got it. Thank you guys.