1- What is the tension in the wire? 2-What is the net force the hinge exerts on
ID: 1409219 • Letter: 1
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