Question

A horizontal beam of length 4.07 m and weight 150 N is attached to a vertical wall at a right angle. There is also a support wire of length 5.01 m which attaches the far end of the beam to the wall from above. Off of the far end of the beam also hangs an object of weight 290 N.

A) What is the magnitude of the tension in the support wire?

B) What is the horizontal component of the force exerted on the beam at the wall?

C) What is the vertical component of the force exerted on the beam at the wall?

Answer 1

let
L = 4.07 m
l = 5.01 m
W1 = 150 N
W2 = 290 N

angle made by the wire with beam, theta = cos^-1(L/l)

= cos^-1(4.07/5.01)

= 35.7 degrees

A) Let T is the tension in the string.

As the beam is in equilibrium net force and net torque acting on it must be zero.

Apply, net torque on the beam = 0

T*L*sin(35.7) - W1*(L/2)*sin(90) - W2*L*sin(90) = 0

T*sin(35.7) = W1/2 + W2

T = (W1/2 + W2)/sin(35.7)

= (150/2 + 290)/sin(35.7)

= 625 N <<<<<<<----------------Answer

B) Let Fx is the horizontal component of the force exerted on the beam at the wall.

Apply, Fnetx = 0

Fx - T*cos(theta) = 0

Fx = T*cos(theta)

= 625*cos(35.7)

= 508 N <<<<<<<----------------Answer

C) Let Fy is the vertical component of the force exerted on the beam at the wall.

Apply, Fnety = 0

Fy + T*sin(theta) - W1 - W2 = 0

Fy = W1 + W2 - T*sin(theta)

= 150 + 290 - 625*sin(35.7)

= 75.2 N <<<<<<<----------------Answer