Sir Lost-a-Lot dons his armor and sets out from the castle on his trusty steed i
ID: 1409386 • Letter: S
Question
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 8.00 m long and has a mass of 1600 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 1100 kg. Hint: First determine all the angles and lengths of the triangle made by the wall, the cable, and the drawbridge.
(a) Determine the tension in the cable
______________
(b) Determine the horizontal force component acting on the bridge at the hinge.
magnitude
________ direction
is it to the left or right?
(c) Determine the vertical force component acting on the bridge at the hinge.
magnitude
_________ direction
is it upwards or downwards
please state the answer clearly with units. thanks
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.