1- What is the tension in the wire? 2-What is the net force the hinge exerts on

Question

1- What is the tension in the wire?

2-What is the net force the hinge exerts on the beam?

3- The maximum tension the wire can have without breaking is T = 1076 N.

What is the maximum mass sign that can be hung from the beam?

4- What else could be done in order to be able to hold a heavier sign?

a-while still keeping it horizontal, attach the wire to the end of the beam

b-keeping the wire attached at the same location on the beam, make the wire perpendicular to the beam

c-attach the sign on the beam closer to the wall

d- shorten the length of the wire attaching the box to the beam

Hanging Beam 1 23 4 5 Go to Question 1 A purple beam is hinged to a wall to hold up a blue sign. The beam has a mass of mb 6.3 kg and the sign has a mass of ms 17.8 kg. The length of the beam is L- 2.67 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 e-30.8.

Explanation / Answer

HORIZONTAL

Fx + T = 0

VERTICAL

Fy - Mb*g - Ms*g = 0

Fy = (Mb+Ms)*g = (6.3+17.8)*9.8 = 236.18 N

TORQUE = 0

T*(2/3)*L*sin30.8 = Mb*g*(L/2)*co30.8 + Ms*g*L*cos30.8

T*(2/3)*2.67*sin30.8 = (6.3*9.8*(2.67/2)*cos30.8) + (17.8*9.8*2.67*cos30.8)

QUESTION 1

T = 516.62 N
Fx = -T = -516.62 N

QUESTION 2

F = sqrt(Fx^2+Fy^2) = 568.04 N

QUESTION 3

T = 1076 N

1076*(2/3)*L*sin30.8 = Mb*g*(L/2)*co30.8 + Ms*g*L*cos30.8

1076*(2/3)*2.67*sin30.8 = (6.3*9.8*(2.67/2)*cos30.8) + (Ms*9.8*2.67*cos30.8)

Ms = 13.20 kg

QUESTION 4

keeping the wire attached at the same location on the beam, make the wire perpendicular to the beam
attach the sign on the beam closer to the wall

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