Sir Lance a Lost’s new draw bridge was designed poorly and stops at an angle of
ID: 250634 • Letter: S
Question
Sir Lance a Lost’s new draw bridge was designed poorly and stops at an angle of 20 degree below the horizontal. Sir Lost and his steed stop when their combined center of mass is 1.0 m from theend of the bridge. The bridge is 8.0 m long and has a mass of 2000 kg; the lift cable is attached to the bridge 5.0 m from the castle end and to a point 12 m above the bridge. Sir Lost’s mass combined with his armor and steed is 1000 kg. Determine (a) the tension in the cable and (b) the horizontal and vertical force components acting on the bridge at the castle end.
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 = 18.9 deg, A = 51.1 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 + 1000 g x 7 sin 70
where T is cable tension
Solving T = 35.5KN
Only horizontal force causing a reaction at the hinge is the horizontal component of the tension (the weights are vertical)
H = T sin B = 35.5*103 sin18.90 =11.5K N
Direction: opposite to horizontal component of T.
Vertical component of tension = T cos A = 35.5*103 cos51.10 =22.3K N
That is not the only vertical force on the bridge, there are the weights to consider. So
V + T cos B - 2000 g - 1000 g = 0
V = -4186 N
Direction: figure out from sin of V, positive upwards..
A = 51.1, B = 18.9, T = 33.5K N, H = 11.5K N, compression, V = -4186 N, downwards.
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