Vector Application Question (worried im wrong)

Question

I'm having trouble with this question.

""Three chains attached to a metal ring are being pulled by different people. Christiane is exerting a force of 1200N at an angle of 30◦ to the horizontal and Hayley is exerting a force of 200N at an angle of 210◦ to the horizontal. What force and in which direction must Benjamin be exerting this force if the ring does not move? Round your answers to one decimal place"

Answer 1

You have done the problem the right way, but made two crucial arithmetic errors: $$1200\cos 30^{\circ} - 200 \cos 30^{\circ} = 1000\cos 30^{\circ} = 500\sqrt{3},$$ and similarly for the other set of equations. (You multiplied $\cos 30^{\circ}$ by itself for some reason.)

If you make those changes and rework the problem, you should get an answer of $1000$ N in the direction of $210^{\circ}$.

There is a much easier way to solve this particular problem, though: just recognize that the Christiane and Hailey are pulling in diametrically opposite directions, so all Benjamin needs to do is to pull in the same direction as Hailey with a force that will balance. Of course, that only works in cases like this where the forces relate to each other in some simple way.

Answer 2

Benjamin's application of force must be equal and opposite to the combined force of Christine and Hayley.

$C = (1200 \cos 30^\circ, 1200\sin 30^\circ) = (600\sqrt 3, 600)\\ H = (200\cos 210^\circ, 200\sin 210^\circ) = (-100\sqrt 3, -100)$

Hayley and Christine are pulling in opposite directions! Ben is pulling in the same direction as Hayley, and pulling with 1000 N force.

If the problem were set up to be a little bit more difficult, we would say.

$H+C = (500\sqrt 3, 500)\\ B + H + C = 0\\ B = -(H+C) = (-500\sqrt 3, -500)$

Magnitude $= \|B\| = \sqrt {(-500\sqrt 3)^2 + (500)^2}$

And the direction is $\text {atan2} \frac{-500}{-500\sqrt 3}$

The atan2 function is similar to the arctan function, except it gives additional directional information when the denominator is negative.

Ben is pulling

Answer 3

Christiane $$C_x=1200\cos{\frac{\pi}{6}}=600\sqrt{3}N$$ $$C_y=1200\sin{\frac{\pi}{6}}=600N$$

Hayley $$H_x=200\cos{\frac{7\pi}{6}}=-100\sqrt{3}N$$ $$H_x=200\sin{\frac{7\pi}{6}}=-100N$$

Since we want the ring not moving at all, then the sum of vectors in $x$-axis and $y$-axis should be equal to 0 $$B_x=-C_x-H_x=-500\sqrt{3}N$$ $$B_y=-C_y-H_y=-500N$$

Because $\tan{\theta}=\frac{B_y}{B_x}=\frac{1}{\sqrt{3}}$, then $\theta=\frac{7\pi}{6}$

Benjamin should pull the chain with the force of $1000N$ at $\frac{7\pi}{6}$ direction.