Question

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 223 N, and Charles pulls with force
of magnitude 189 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?

Answer 1

Answer:

The magnitude of Betty's force

if Charles pulls in (a) the direction drawn in the picture is 0 and (b) the other possible direction for equilibrium is Fbetty = 358.29 N

Explanation:

Let Betty's angle is 90°

And suppose the angle of Alex= 153° from the -y axis or it may be shown as 153-90= 63° from -x axis.

Let Charles's angle be θ

Net force is zeo. Therefore, Falex+Fbetty+Fcharles=0

Let y component be Fybetty=-Fyalex-Fycharles

Fb * sin(-90) = - Fa * sin(117) - Fc * sin(θ)

This can't be calculated.

Let x component be as Fbettyx= -Fzalex-Fcharlesx

Fb * cos (-90) = - Fa * cos(117) - Fc * cos(θ)

0 = -223 * cos(117) - 189 * cos(θ)

θ = cos-1 (-210*cos(117) / 187)

θ=57.61

Now, Fbettyy * sin(-90) = - Fex * sin(117) - Fcharles * sin(57.61)

- Fbetty = -223 * sin(117) - 189 * sin(57.61)

Fbetty=358.29