Sir Lance a Lost new draw bridge was designed poorly and stops at an angle of 20o below the horizontal. Sir Lost and his steed s
top when their combined center of mass is 1.0 m from the end of the bridge. The bridge is 8.0 m long and has a mass of 2000 kg; the lift cable is attached to the bridge 5.0 m from the castle end and to a point 12 m above the bridge. Sir Lost’s mass combined with his armor and steed is 1000 kg.
Determine
(a) the tension in the cable and
(b) the horizontal and vertical force components acting on the bridge at the castle end.
Determine
(a) the tension in the cable and
(b) the horizontal and vertical force components acting on the bridge at the castle end.
1 answer:
Answer:
The Tension T is 42120N
The Horizontal force component is 18322.2N
The Vertical force component is - 4729N
Explanation:
First, you have to find the angle between the drawbridge and the cable using sine and cosine rule. This will result in angle 44.2°. Hence, the angle between the horizontal axis and the cable will be 64.2° (44.2° + 20°).
Having done that, you apply two conditions of equilibrium.
1. THE VECTOR SUM OF ALL FORCES EQUAL ZERO.
∑Fx = 0
∑Fx = Rx - Tcos64.2 = 0
Rx = 0.435T
∑Fy = 0
∑Fy = Ry + Tsin64.2 - W - w = 0
W = 2000kg × 9.8 = 19600N
w =1000kg × 9.8 = 9800N
Ry + 0.9T = 29400N
Ry = 29400 - 0.9T
2. THE SUM TOTAL OF TORQUES EQUALS ZERO
Rx: τ = 0
Ry: τ = 0
T: τ = 5 × Tsin44.2
= 3.49T m
W: τ = 4 × 19600sin90
= 78400Nm
w: τ = 7 × 9800sin9
= 68600Nm
Note:
Rx = x component of Reaction force
Ry = y component of Reaction force.
T = Tension
W = weight of bridge
w = weight of Sir Lance a Lost and his steed
τ = torque
Note: The torque of Tension is counter clockwise while that of the weights is clockwise.
Hence,
∑τccw = ∑τcw
3.49T = 78400 + 68600
3.49T = 14700Nm
T = 147000/3.49
T = 42120N
Rx = 0.435 × 42120
Rx = 18322.2N
Ry = 29400N - (0.9×42120)N
Ry = 29400 - 34129
Ry = -4729N
Note: Ry being negative means that the hinge of the drawbridge exerts a downward force.
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