How Does Sir Lost-a-Lot's Quest Impact the Forces on the Drawbridge?

[eightbrown8]

Sir Lost-a-Lot dons his armor and sets out from the castle on his trusty steed in his quest to improve communication between damsels and dragons. Unfortunately his squire lowered the draw bridge too far and finally stopped it 20.0° below the horizontal. Lost-a-Lot and his horse stop when their combined center of mass is 1.00 m from the end of the bridge. The uniform bridge is 10.00 m long and has a mass of 1800 kg. The lift cable is attached to the bridge 5.00 m from the hinge at the castle end, and to a point on the castle wall 12.0 m above the bridge. Lost-a-Lot's mass combined with his armor and steed is 950 kg.

(a) Determine the tension in the cable.
kN

(b) Determine the horizontal force component acting on the bridge at the hinge.
magnitude
kN
direction
to the left
to the right


(c) Determine the vertical force component acting on the bridge at the hinge.
magnitude
kN
direction
downwards
upwards

 

[Astronuc]

This is a statics problem so that the sum of the forces in horizontal (x) and vertical (y) directions are zero. Furthermore, the net moment of the bridge is zero.

The center of mass (weight) of the knight and horse act downward at 9 m from the pivot. The mass of the bridge acts downard at it center of mass, midway along the bridge. The rope/cable acts at some angle with respect to horizontal and that angle must be determined. The cable is attached at the midpoint of the bridge. Unfortunately the triangle formed by bridge, wall and cable is not a right triangle, but rather the largest angle is 110°.

Please make an effort at solving the problem and show the work.

 

[ogb p]

(a) To determine the tension in the cable, we can use the principle of static equilibrium, which states that the sum of all forces acting on an object must be equal to zero. In this case, we can consider the bridge as the object of interest. The forces acting on the bridge are its weight (mg), the tension in the cable (T), and the reaction force at the hinge (R). We can set up the following equation:

ΣF = 0
mg + T + R = 0

Since we know the values of m (1800 kg) and g (9.8 m/s²), we can rearrange the equation to solve for T:

T = -(mg + R)

To determine the value of R, we can use the principle of moments, which states that the sum of all moments acting on an object must be equal to zero. In this case, we can choose the hinge as the point of interest. The forces acting on the bridge that create a moment are the weight of the bridge (mg) and the tension in the cable (T). We can set up the following equation:

ΣM = 0
mg(5.00 m) + T(12.0 m) = 0

Solving for T, we get:

T = -(mg(5.00 m)) / 12.0 m
T = -1500.0 N

Finally, we can substitute this value of T into the equation we found earlier to solve for the tension in the cable:

T = -(mg + R)
-1500.0 N = -(1800 kg)(9.8 m/s²) + R
R = 33600 N

Therefore, the tension in the cable is 1500.0 N and the horizontal force component acting on the bridge at the hinge is 33600 N to the left.

(b) To determine the horizontal force component acting on the bridge at the hinge, we can use the same equation we found earlier for the tension in the cable:

T = -(mg + R)

Since we know the value of T (1500.0 N), we can solve for R:

R = -(mg + T)
R = -(1800 kg)(9.8 m/s² + 1500.0 N)
R = 33600 N

Therefore, the horizontal force component acting on the bridge at the hinge is 33600 N to