magnitude N direction Sir Lost-a-Lot dons his armor and sets out from the castle
ID: 2260152 • Letter: M
Question
magnitude N directionSir 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 I¸ = 20.0A^ degree 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 scripted l = 7.00 m long and has a mass of 2 400 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 1 200 kg.
Explanation / Answer
First thing first, that is Geometry. We need to know the angle the cable makes with the vertical (and the bridge), in order to find its magnitude and components.
Let's designate some points: A is where the cable attaches to the bridge, O the hinge of the bridge, B where the cable attaches to the castle wall. We have, in the triangle AOB, OB = 12 m, OA = 5 m.
Now, OB is vertical and OA is 20 deg below horizontal, that makes angle AOB 110 deg, hence, angles OAB + OBA (call them A and B) is supplement to 110 deg:
A + B = 70 or A = 70 - B
Also, applying Sine Rule of Triangles:
sinA / 12 = sinB / 5
12 sinB = 5 sin(70 - B) = 5 sin 70 cosB - 5 cos70 sinB
tanB = ...
Solving, B = ... deg, A = ... deg
Now that Geometry is out of the way, let's delve into Statics. Moment of a force about a point is the cross (outer) product of the force and radius vector (line from point of moment to point of action of force). If we take moment about the hinge, the moment of the cable tension must balance the moments of the weights of the bridge and the knight and stead.
Also if the angle between the bridge and horizontal is 20 deg, then angle between bridge and vertical (direction of weight) must be 70 deg. So here we go
T x 5 sin 51.1 = 2000 g x 4 sin 70 + 1200 g x 7 sin 70
where T is cable tension
Solving T = ... N
Only horizontal force causing a reaction at the hinge is the horizontal component of the tension (the weights are vertical)
H = T sin B = ... N
Direction: opposite to horizontal component of T.
Vertical component of tension = T cos A = ... N
That is not the only vertical force on the bridge, there are the weights to consider. So
V + T cos B - 2000 g - 1200 g = 0
V = ... N
Direction: figure out from sin of V, positive upwards..
A = 51.1, B = 28.9, T = 38852 N, H = 12585 N, compression, V = -5365 N, downwards.
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