Question

A window washer is standing on a plank supported by a vertical rope at each end. The plank weighs 201 N and is 2.90 m long. What is the tension in each rope when the 707 N worker stands 1.03 m from one end?

Answer 1

Step-by-step explanation:

Given:

w₁ = 707 N

w₂ = 201 N

r₁ = 1.03 m

r₂ = 1/2 × 2.90 m = 1.45 m

[1]

∑Fy = 0

T₁ + T₂ - w₁ - w₂ = 0

T₁ + T₂ = w₁ + w₂

T₁ + T₂ = 707 + 201

T₁ + T₂ = 908

[2]

Στ = 0

T₁(0) + w₁r₁ + w₂r₂ + T₂(2.90) = 0

0 + 707(-1.03) + 201(-1.45) + 2.9T₂ = 0

T₂ = 1019.66 ÷ 2.9

T₂ = 351.61 N

T₁ + T₂ = 908

T₁ + 351.61 = 908

T₁ = 556.39 N

Answer 2

To calculate the tension in each rope supporting a plank with a window washer on it, we use the principles of static equilibrium, setting up equations for the sum of vertical forces and the sum of torques to solve for the tension in each rope.

Step-by-step explanation:

The subject of the question is Physics, specifically related to static equilibrium and the calculation of tension in cables supporting loads. To find the tension in each rope when a 707 N worker stands 1.03 m from one end of a 2.90 m long plank that weighs 201 N, we use the principles of torques in equilibrium. The sum of forces and the sum of torques (moments) around any point must be zero for the system to be in static equilibrium.

Let's choose the left end of the plank as the pivot point. The worker applies a downward force of 707 N at a distance of 1.03 m from the pivot. The plank applies a downward force of 201 N at its center of gravity, which is 1.45 m (half of the plank's length) from the pivot. If we designate T1 as the tension in the left rope and T2 as the tension in the right rope, we can write two equations:

1. The sum of vertical forces is zero: T1 + T2 = weight of worker + weight of plank.
2. The sum of torques around the pivot is zero: T2 × 2.90 m = (weight of worker × 1.03 m) + (weight of plank × 1.45 m).

By solving these two equations simultaneously, we can find the tension in each rope supporting the plank and worker.