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