In a two-dimensional tug-of-war, Alex, Betty, and Charles pull horizontally on an automobile tire at the angles shown in the picture. The tire remains stationary in spite of the three pulls. Alex pulls with force of magnitude 207 N, and Charles pulls with force of magnitude 188 N. Note that the direction of is not given. What is the magnitude of Betty's force if Charles pulls in (a) the direction drawn in the picture or (b) the other possible direction for equilibrium?

Question

I already found the correct answer for a to be 291 N, but I am having trouble finding b.

If you copy and paste this question in the Google search bar, the exact same diagram is shown in the problem but with the angle 140 degrees. There are many results with this similar question. Just copy and paste this question and click the first link that appears on the first page (I cannot type the url)and there are three problems on the page and the image to this problem is on the bottom.

Bot · Answer

For part b, Charles will be pulling in the opposite direction. First, let's find the x and y components of Charles' pull:

Fx_charles = 188 * cos(140) = -144.43 N
Fy_charles = 188 * sin(140) = 120.25 N

Now, we already know Alex's pull and the components:

Fx_alex = 207 * cos(100) = -88.84 N
Fy_alex = 207 * sin(100) = 193.8 N

For the tire to be in equilibrium, the sum of the x-components of the forces must be equal to zero. Similarly, for the sum of the y-components.

So, Fx_betty - 144.43 - 88.84 = 0 => Fx_betty = 233.27 N
And, Fy_betty + 120.25 - 193.8 = 0 => Fy_betty = 73.55 N

Now, to find the total magnitude of Betty's force, we can use the Pythagorean theorem:

magnitude = sqrt(Fx_betty^2 + Fy_betty^2) = sqrt(233.27^2 + 73.55^2) ≈ 244.59 N

So, the magnitude of Betty's force in part b is approximately 244.59 N.