Forces on a hinged board used as a lever

[ccitycoop]

I'm having trouble with this one. Any help would be appreciated.

A 3 m board of mass 4.5 kg is hinged at one end. A force F is applied vertically at the other end to lift a 75 kg block, which rests on the board 80 cm from the hinge, as shown in Figure 12-35.

(b) Find the force exerted by the hinge at this angle. (in components)
(c) Find the magnitude of the force F if this force is exerted perpendicular to the board when = 30°.
(d) Find the force exerted by the hinge in this case (in components)

 

[Hootenanny]

Welcome to the Forums,

One is expected to show ones own efforts before requesting help.

 

[kyle8921]

I would first clarify the setup and terminology used in the question. The board is being used as a lever, with the hinge acting as the fulcrum. The force F is being applied at the opposite end of the board to lift a block, which is positioned 80 cm from the hinge. The angle at which the force F is being applied is denoted as θ.

To solve part (b), we can use the principle of moments, which states that the sum of the moments (or torques) acting on a system must be equal to zero for the system to be in equilibrium. In this case, the hinge is acting as the fulcrum and the weight of the board and block are acting downwards, creating a clockwise moment. The force F is acting upwards, creating a counterclockwise moment. Therefore, we can set up the following equation:

Mhinge + Mboard + Mblock = 0

Since the board is hinged at one end, the moment of the hinge (Mhinge) is equal to zero. We can also assume that the weight of the board is negligible compared to the weight of the block, so the moment of the board (Mboard) can also be ignored. This leaves us with:

Mblock = F * d

Where d is the distance between the force F and the hinge, which in this case is 3 m – 0.8 m = 2.2 m. The moment of the block (Mblock) can be calculated as:

Mblock = (75 kg * 9.8 m/s^2) * 0.8 m = 588 Nm

Setting this equal to the moment of the force F, we can solve for the force F:

588 Nm = F * 2.2 m
F = 588 Nm / 2.2 m = 267.27 N

To find the force exerted by the hinge, we can use the law of cosines to find the magnitude of the force (FH) and the law of sines to find the angle (α) between the force and the horizontal:

FH^2 = F^2 + (75 kg * 9.8 m/s^2)^2 – 2 * F * 75 kg * 9.8 m/s^2 * cos(150°)
FH = 539.5 N

sin(α) = F