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A stationary 3.0-m board of mass 5.5-kg is hinged at one end. A force is applied vertically at the other end, and the board makes a 30° angle with the horizontal. A 55-kg block rests on the board 80 cm from the hinge as shown in the figure below.

(a) Find the magnitude of the force F.

(b) Find the force exerted by the hinge.

(c) Find the magnitude of the force F as well as the force exerted by the hinge (F_h), if F is exerted, instead, at right angles to the board.

Added Sat, 06 Jun '15

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What are our givens

theta = 30°

M = 5.5kg = mass of Board

m = 55kg = mass of Block

L = 3 meters = length of board

x = 0.8 meters - distance of block from hinge

(a) Find the magnitude of the force F.

First express our static equilibrium equations.

Forces in the Y direction = Fy

Fy = F + F_h - Mg - mg = 0

>> F + F_h = Mg + mg

There are no forces in the X direction

Torque as defined from the reference point being at the hinge.

Tq = L*F*cos(30) - x * mg*cos(30) - L/2 * Mg * cos(30) = 0

>> L*F - x*mg - L/2*Mg = 0

Now solve for F

L*F = x*mg + L/2 * Mg

>> F = x / L * mg + 1/2 *Mg

F = 0.8-m / 3-m * (55-kg) * 9.81-m/s^2 + 1/2 * 5.5-kg * 9.81-m/s^2

F = 170.86 N

(b) Find the force exerted by the hinge.

Use F + F_h = Mg + mg and plug in F

F_h = Mg + mg - F

F_h = g(M+m) -F

F_h = 9.81-m/s^2(5.5-kg + 55-kg) - 170.86 N

F_h = 422.65

(c) Find the above if F is at right angles to the board.

This will change both of our equations a bit by adding a horizontal force component as well as making it impossible to divide out the cos(30) in the torque equation.

Fy = F + F_hy - Mg - mg = 0

>> F*cos(30) + F_hy = Mg + mg

Fx = F*sin(30) - F_hx = 0

>>F *sin(30) = F_hx

Torque as defined from the reference point being at the hinge.

Tq = L*F - x * mg*cos(30) - L/2 * Mg*cos(30) = 0

>> F = x/L * mg * cos(30) + 1/2 * Mg * cos(30)

Solve for F

F = 0.8-m / 3-m * 55-kg * 9.81-m/s^2*cos(30) + 1/2 * 5.5-kg * 9.81-m/s^2 * cos(30)

F = 147.96-N

Now solve for F_h (F_h = sqrt((F_hx)^2 + (F_hy)^2))

F*sin(30) = F_hx

F_hx = 147.96-N * sin(30)

F_hx = 73.98-N

F*cos(30) + F_hy = Mg + mg

>> F_hy = g(M + m) - F*cos(30)

F_hy = 9.81-m/s^2(5.5-kg + 55-kg) - 147.96-N * cos(30)

F_hy = 465.36

F_h = 471.21

Added Sat, 06 Jun '15

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