Calculating Tension of Rope for Scaffold with 690N Worker

  • Thread starter Thread starter esinn08
  • Start date Start date
  • Tags Tags
    Rope Tension
AI Thread Summary
To calculate the tension in the ropes supporting a scaffold with a 690N worker and a 190N scaffold, the sum of forces and torques must equal zero. The scaffold's weight acts at its midpoint, while the worker's position affects the tension distribution. By summing torques about one end and setting them equal to zero, one can solve for the tensions in the ropes. The equations must account for the distances from the pivot point to each force. Properly applying these principles will yield the correct tension values in each rope.
esinn08
Messages
8
Reaction score
0
Hi everyone,

My question is as follows:

A window washer is standing on a scaffold supported by a vertical rope at each end. The scaffold weighs 190N and is 2.9m long. What is the tension in each rope when the 690N worker stands 1.9m from one end?

I know the sum of the forces has to be 0, since the scaffold isn't moving. I also know to get the tension you multiply the weight by the appropriate length. I'm just not sure where to go from there. Any suggestions would be greatly appreciated :smile:

Thanks for your help!
 
Physics news on Phys.org
esinn08 said:
Hi everyone,

My question is as follows:

A window washer is standing on a scaffold supported by a vertical rope at each end. The scaffold weighs 190N and is 2.9m long. What is the tension in each rope when the 690N worker stands 1.9m from one end?

I know the sum of the forces has to be 0, since the scaffold isn't moving. I also know to get the tension you multiply the weight by the appropriate length. I'm just not sure where to go from there. Any suggestions would be greatly appreciated :smile:

Thanks for your help!
The sum of the forces AND the sum of the torques must equal zero. The scaffold weight may be represented by a single 190N force acting 1.45m from one end (that is, in the middle). The tension forces are unknown at this point, call them T_1 and T_2 (they will act UP). By summing torques (force times perpendicular distance) about one end, and setting them equal to zero, watching your plus and minus signs (clockwise is plus and counterclockwise is minus), you can solve for one of the tension forces, and then the other by noting sum of all forces in vertical direction is 0.
 
(I made a typo in my question, the scaffold is 690N and the person is 190N.)

Regardless, I am not sure that I am completely understanding what you wrote.

(1.45m)(-690N) + (T_1)(1.45m) = 0

(1.9m)(-190N) + (T_2)(1.0m) = 0

I have a feeling that is not right. Any more suggestions?
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top