Answer:
The weight of the bridge acts at its midpoint, which is 2 m from either support. Therefore, each support will bear half the weight of the bridge, which is 20 N / 2 = 10 N.
The weight of the toy car acts at a distance of 1 m from support A. Therefore, the weight of the toy car will produce a clockwise moment around support B, given by:
M = Fd = (4 N)(3 m) = 12 Nm
where F is the weight of the toy car and d is the distance between the weight and support B.
This moment must be balanced by a counterclockwise moment produced by the support forces. The counterclockwise moment around support B is given by:
M = F2d2 = (10 N)(2 m) = 20 Nm
where F2 is the support force at support B and d2 is the distance between support B and the weight of the bridge.
Therefore, we have:
F2 = M / d2 = 20 Nm / 2 m = 10 N
The vertical forces must also balance. The total downward force is the weight of the bridge plus the weight of the toy car, which is 20 N + 4 N = 24 N. This force must be balanced by the upward support forces, which are F1 and F2. Therefore, we have:
F1 + F2 = 24 N
Substituting the value of F2 we just found, we get:
F1 + 10 N = 24 N
Solving for F1, we get:
F1 = 14 N
Therefore, the support forces are F1 = 14 N and F2 = 10 N.