Calculating Tension in Ropes: Window Washer on Suspended Plank - 20kg Mass

[bigsaucy]

Hello all, just another book problem that i need clarification with. Thanks in advance

1) A window washer works on a plank of mass 20kg suspended by ropes from the roof of the building. The plank is a uniform piece of wook 3.0 meters wide. The window washer has a mass of 75kg and stands 0.5m from the left side of the plank.

Calculate the tensions in the ropes holding up the plank of wood.

My Answer: Tension of rope on left side (Closer to worker) is 661.94N
Tension of rope on right side (further away from worker) is 269.06N

 

[da_nang]

You need to show us your work so we can see where you've made mistakes, as I get different answers.

 

[bigsaucy]

hello da_nang

I started off by reasoning that the centre of gravity of the plank was at 1.5m (dead centre of the plank) but when the 75kg man stood on the plank the centre of gravity shifts to the left slightly. I used the postion of the man at 0.5 meters from the left as the reference point for calculating the centre of gravity. Therefore I got the two points (0,75) and (1.5,20). With the (1.5,20) being c.o.g of the plank and it's mass. I then calculated the new c.o.g of the plank + the man to be 0.711m from the left of the plank.

I then reasoned that because gravity would pull down on the c.o.g if the left string was not there, that it had to be exerting a torque on the plank which i found from the torque equation rF to be (0.711m)(931N) = 661.9N.

I then used this equation T(1) + T(2) - mg = 0 because the system is in equilibrium, to find the tension of the string on the right.

 

[gneill]

I then reasoned that because gravity would pull down on the c.o.g if the left string was not there, that it had to be exerting a torque on the plank which i found from the torque equation rF to be (0.711m)(931N) = 661.9N.

That's a torque, not a tension. The units should be Nm. (Newton-meters).

In order to find the tension that this torque produces at the far end of the plank you must set up the appropriate equation that balances this torque with the torque caused by the tension in the rope. If the tension in the far end rope is T₂, the weight of the combined plank and man w, the distance of the center of gravity from the left end is d1, and the total length of the plank is L, then

d1*w = L*T2

 

[da_nang]

I then reasoned that because gravity would pull down on the c.o.g if the left string was not there, that it had to be exerting a torque on the plank which i found from the torque equation rF to be (0.711m)(931N) = 661.9Nm.

(The bolded part is a correction) Correct, that's the torque that the gravity exerts on the plank if the center of rotation is at the point where the left rope connects to the plank.

T₁ is the tension in the left rope.
T₂ is the tension in the right rope.

Then if you use this value in the net torque equation T₂*r₂ - G*r = 0 you should get one of the tension forces. You then use this value and the gravity force in Newton's second to get the other tension force.