Sir Lost-a-Lot dons his armor and bursts out
of the castle on his trusty steed Tripper in
his quest to rescue fair damsels from drag-
ons. Unfortunately his valiant aide Doubtless
lowered the drawbridge too far and finally
stopped it 30� below the horizontal. Lost-
a-Lot screeches to a stop when his and his
steed’s combined mass (1500 kg) is 1.2 m
from the end of the bridge, which is 7.5 m long with a mass of 2300 kg. The lift cable is
attached to the bridge 3.3 m from the hinge
and to the parapets 9.7 m above the bridge.
Find the tension in the cable. The acceler-
ation of gravity is 9.8 m/s2 .
this is so difficult please help
To find the tension in the cable, we can use the principle of moments, which states that the sum of the anticlockwise moments is equal to the sum of the clockwise moments.
First, let's consider the bridge as our pivot point. The anticlockwise moments include the weight of Lost-a-Lot and his steed (1500 kg × 9.8 m/s²), acting downward at a distance of 1.2 m from the end of the bridge. The clockwise moments include the weight of the bridge itself (2300 kg × 9.8 m/s²), acting downward at a distance of (7.5 m / 2) = 3.75 m from the pivot point.
The tension in the cable creates a clockwise moment by lifting the bridge. It acts at a vertical distance of 9.7 m above the bridge. The horizontal distance between the pivot point and the cable attachment is 3.3 m.
Now, let's set up the equation based on the principle of moments:
(1500 kg × 9.8 m/s² × 1.2 m) = (2300 kg × 9.8 m/s² × 3.75 m) + (tension × 9.7 m × 3.3 m)
Simplifying this equation gives:
17640 = 84525 + 31.71 × tension
Rearranging the equation, we have:
31.71 × tension = -66885
Dividing both sides of the equation by 31.71, we find:
tension = -2107.66 N
Since tension cannot be negative, this means that the direction of force exerted by the cable is opposite to the assumed direction. Therefore, the tension in the cable is approximately 2107.66 N.