a 200 N wooden platform is suspended from the roof of the house by ropes attached to it's ends. it has a total length of 4.2 meters. a painter weighing 650 N stands from the left end of the platform. find the tension in each end of the ropes .
how far from the end of the platform?
To find the tension in each end of the ropes, we need to analyze the forces acting on the suspended platform.
Let's denote the tension in the left rope as T1 and the tension in the right rope as T2.
1. First, consider the forces acting vertically:
- The weight of the wooden platform is 200 N, acting downward.
- The weight of the painter is 650 N, also acting downward.
- The total downward force is the sum of the weight of the platform and the painter, which is 200 N + 650 N = 850 N.
2. Second, consider the forces acting horizontally:
- There are no forces acting horizontally since the platform is not moving in that direction.
Since the platform is at equilibrium (not accelerating in any direction), the sum of the forces in both the vertical and horizontal directions must be zero.
Let's focus on the vertical forces first:
- The tension in the left rope, T1, is pulling upward.
- The tension in the right rope, T2, is also pulling upward.
- The weight of the platform and the painter, 850 N, is pulling downward.
Based on the equilibrium condition, the sum of the vertical forces is zero:
T1 + T2 - 850 N = 0
To find the tension in each rope, we can consider the horizontal forces (which are zero in this case):
- The tension in the left rope, T1, is pulling to the right.
- The tension in the right rope, T2, is pulling to the left.
Since there are no horizontal forces, T1 and T2 must be equal in magnitude:
T1 = T2
Substituting T1 = T2 into the equation we obtained from the vertical forces:
T1 + T1 - 850 N = 0
2T1 - 850 N = 0
2T1 = 850 N
T1 = 425 N
Therefore, the tension in each end of the ropes is 425 N.