Question

A uniform beam 8 m long is supported by a horizontal cable (at the top) and a hinge as shown in the picture. A 35 N box is attached at the end of the beam. The tension in the horizontal cable is 400 N. a) Find the weight of the beam. b) Find the force of the hinge on the Beam. c) The 35 N box is removed but in the process, the cable is cut by mistake. Find the angular acceleration of the beam immediately thereafter.

Please include the formulas you used to solve the problem. Thank you.

Answer 1

a)

here , let the weight of the beam is W

balancing the moment of forces about th ehinge

35 * 8 * sin(50) - 400 * 8 * cos(50) + W * (8/2) * sin(50) = 0

solving for W

W = 601.3 N

the weight of the rod is 601.3 N

b)

at the hinge ,

Fx = 400 N

Fy = 601.3 + 35

Fy = 636.3 N

Now , net force = sqrt(Fx^2 + Fy^2)

net force = sqrt(636.3^2 + 400^2)

net force = 751.84 N

the force on the hinge is 751.84 N

c)

let the angular accelration is a

m*l^2/3 * a = m * g * l/2 * sin(50)

l/3 * a = g * sin(50)/2

a = 9.8 * sin(50) * 3/(2 * 8)

a = 1.41 m/s^2

the angular acceleration is 1.41 m/s^2