Question

At a picnic, there is a contest in which hoses are used to shoot water at a beach ball from three directions. As a result, three forces act on the ball, F1, F2, and F3 (see drawing). The magitudes of 1 and 2 are F1 = 55.0 newtons and F2 = 95.0 newtons. Using a scale drawing and the graphical technique determine the following such that the resultant force acting on the ball is zero.

i'm not sure how to start this problem
help pleasee

Answer 1

Figure out the East-West forces and add them up:
... from F1: 40 cos(60 deg) = 20
... from F2: -90
... from F3: F3 cos(theta)
Add them up:
... -70 + F3 cos(theta)
Set it equal to zero (the ball does not move)
... 70 = F3 cos(theta)

Now do the same for the North-South forces. You should get
... 34.641 = F3 sin(theta)

We have two equations in two unknowns. We can solve them, first for the angle:

34.641 / 70 = [-F3 sin(theta) ] / [F3 cos(theta) ]
0.4949 = -tan(theta)
Using arctangent, we find
theta = 26.33 degrees.
And cos(theta) = 0.896
and sin(theta) = 0.444

Putting these back into the first two equations:
70 = F3 cos(theta)
70 / 0.896 = F3 = 78.102
And also
34.641 = F3 sin(theta)
34.641 / 0.444 = F3 = 78.102

Answer 2

the answer above is for another calculation so plug in you values and derive your answer