Question

Sir Lost-a-Lot dons his armor and sets out from the castle on his trusty steed (see figure below). Usually, the drawbridge is lowed to a horizontal position so that the end of the bridge rests on the stone ledge. Unfortunately, Lost-a-Lot's squire didn't lower the drawbridge far enough and stopped it at θ = 20.0° above the horizontal. The knight and his horse stop when their combined center of mass is d = 1.00 m from the end of the bridge. The uniform bridge is script i = 7.50 m long and has a mass of 2 100 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 h = 12.0 m above the bridge. Lost-a-Lot's mass combined with his armor and steed is 900 kg.

(a) Determine the tension in the cable.

__________N

(b) Determine the horizontal force component acting on the bridge at the hinge.

magnitude ________N

(c) Determine the vertical force component acting on the bridge at the hinge.

magnitude ________N

Answer 1

In the given problem relative to the hinge end of the bridge the cable is attached horizontally out a distance \(x=(5.00 \mathrm{~m}) \cos 20.0^{\circ}\)

\(=4.70 \mathrm{~m}\)

Vertically down at a distance of \(y=(5.00 \mathrm{~m}) \sin 20.0^{\circ}\)

\(=1.71 \mathrm{~m}\)

Then the angle made bythe cable with the horizontal will be \(\theta=\tan ^{-1}\left[\frac{(12.0-1.71) \mathrm{m}}{4.70 \mathrm{~m}}\right]=65.45^{\circ}\)

(a) Taking the torques about the hinge end of the bridge we get \(R_{x}(0)+R(0)-20.58 \mathrm{kV}(3.75 \mathrm{~m}) \cos 20.0^{\circ}-\mathrm{I} \cos 65.45^{\circ}(1.7 \mathrm{Im})\)

\(+T \sin 65.45^{\circ}(4.70 \mathrm{~m})-8.82 \mathrm{kN}(6.50 \mathrm{~m}) \cos 20.0^{\circ}=0\)

So this yields to get the tension T in the cable as \(\mathrm{T}=\mathrm{B} \cdot 235 \mathrm{kN}\)

(b) In the \(x\) - direction \(\sum F_{x}=0\)

\(R,-T \cos 71.1^{\circ}=0\)

\(R_{x}=(5.235 \mathrm{kN}) \cos 65.45^{\circ}\)

So the horizontal component will be \(R_{\mu}=2.172 \mathrm{kN}\) (towords right)

(c) In the \(y-\) direction

$$ \begin{array}{l} \sum F_{2}=0 \\ R_{y}=-20.58 \mathrm{kN}+\mathrm{T} \sin 65.45^{\circ}-8.82 \mathrm{kN}=0 \end{array} $$

So the vertical component of force will be

$$ \begin{aligned} R_{x} &=29.40 \mathrm{kN}-(5.32 \mathrm{kN}) \sin 65.45^{\circ} \\ &=24.64 \mathrm{kN} \end{aligned} $$