Math 265
Sections U2, V2, W2
8/25/16
12.2.46 Consider a 50 N weight suspended by two wires as shown below. If the magnitude of F1 is 35 N, find the angle α and the magnitude of F2 .
Solution: We will break up the forces into components as shown in the picture below.
Notice that the x component of F1 is negative because it pulls in the opposite direction.
Because the weight is in static equilibrium, we know there is no net force on the object (or
equivalently, the forces cancel out.) We can write the following equations
|F2 | · cos(60°) − 35 · cos(α) = 0
|F2 | · sin(60°) + 35 · sin(α) = 50
We know the sine and cosine of 60°, so we’ll write those in right away.
1
|F2 | − 35 · cos(α) = 0
2
√
3
|F2 | + 35 · sin(α) = 50
2
We can solve the first equation for |F2 |:
|F2 | = 70 cos(α)
and substitute it the second.
1
2
√
√
3
|F2 | + 35 · sin(α) = 50
2
3
70 cos(α) + 35 · sin(α) = 50
2
√
1
5
3
cos(α) + sin(α) =
2
2
7
We’re going to substitute back for sine and cosine of 60 (so we may have been better off
just keeping them that way originally):
5
7
and from here we’ll use the identity sin(β + α) = sin(β) cos(α) + cos(β) sin(α), where
sin(60°) cos(α) + cos(60°) cos(α) =
β = 60°.
5
7
and solving for α yields α = −14.415°. This makes sense with the reference frame, since
sin(α + 60°) =
the angle is measured below the support. We know |F2 | = 70 cos(α), so |F2 | = 67.796 N.