A purple beam is hinged to a wall to hold up a blue sign. The beam has a mass of
ID: 1553767 • Letter: A
Question
A purple beam is hinged to a wall to hold up a blue sign. The beam has a mass of mb = 6.2 kg and the sign has a mass of ms = 17.1 kg. The length of the beam is L = 2.5 m. The sign is attached at the very end of the beam, but the horizontal wire holding up the beam is attached 2/3 of the way to the end of the beam. The angle the wire makes with the beam is = 33.5°.
1)
What is the tension in the wire?
N
2)
What is the net force the hinge exerts on the beam?
N
3)
The maximum tension the wire can have without breaking is T = 858 N.
What is the maximum mass sign that can be hung from the beam?
kg
Explanation / Answer
Since the system is in equilibrium position
Net torque acting is zero
Torque due to weight of the bar + Torque due to Blue sign + Torque due to tension in horizontal wire = 0
(ms*g*L*sin(90-theta)) + (m_bar*g*(L/2)*sin(90-theta)) - (T*(2L/3)*sin(theta)) = 0
(17.1*9.8*2.5*sin(90-33.5)) + (6.2*9.8*(2.5/2)*sin(90-33.5)) - (T*(2*2.5/3)*sin(33.5)) =0
T = 448.62 N
2) Horizontal component of Force exerted by hinges is Fx = T = 448.62 N
Vertical force exerted by hinges is Fy = (17.1+6.2)*9.81 = 228.573 N
F = sqrt(Fx^2+Fy^2) = sqrt(228.573^2+448.62^2) = 503.5 N
3) Using Net torque actiong on the system is zero
orque due to weight of the bar + Torque due to Blue sign + Torque due to tension in horizontal wire = 0
(ms*g*L*sin(90-theta)) + (m_bar*g*(L/2)*sin(90-theta)) - (T*(2L/3)*sin(theta)) = 0
(ms*9.8*2.5*sin(90-33.5)) + (6.2*9.8*(2.5/2)*sin(90-33.5)) - (858*(2*2.5/3)*sin(33.5)) =0
ms = 35.53 Kg
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