Be careful with all the angles.

The first condition of equilibrium is that the vector sum of all the forces equals zero.

Fx = 0

Fx = Rx – T cos 64.2 o = 0

Rx = T cos 64.2 o = 0.435 T

Fy = 0

Fy = Ry + T sin 64.2 o – W – w = 0

W = (2 000 kg) (9.8 m/s2) = 19 600 N

w = (1 000 kg) (9.8 m/s2) = 9 800 N

Ry + 0.900 T sin 64.2 o = 29 400 N

As we should expect by now, at this stage, we have three unknowns — Rx, Ry, and T — but only two equations. We will get the third equation from applying the second condition of equilibrium, that the sum of the torques must equal zero. We will calculate the torques about the hinge of the drawbridge.

Rx: = 0

Ry: = 0

T: = ccw = (5 m) T sin 44.2o = (5 m) (T) (0.697) =

= 3.49 m T

W: = cw = (4 m) (19 600 N) sin 90o = 78 400 N-m

w: = cw = (7 m) (9 800 N) sin 90o = 68 600 N-m

ccw = cw

3.49 m T = 147 000 N-m

T = 42 120 N

Now that the tension is known, we can go back and determine the “reaction force” Rx and Ry,

Rx = 0.435 T

Rx =18 322 N

Ry = 29 400 N – 0.900 T sin 64.2 o

Ry = 29 400 N – 34 129 N

Ry = – 4 729 N