a uniform weightless rod 8 meters long weighing 800N is supported horizontaly by two ropes hung 2 meters from its ends.Find the force in each rope when a painter weighing 700N stands from one ends.
How can the "weightless" rod weigh 800 N?
To find the force in each rope, we can use the principle of moments, also known as the law of balance or torque.
The principle of moments states that for an object to be in equilibrium, the sum of the clockwise moments about any point must be equal to the sum of the anticlockwise moments about the same point.
First, let's define the variables we will use in our calculations:
- F1: Force in the first rope
- F2: Force in the second rope
- L: Length of the rod (8 meters)
- W: Weight of the rod (800N)
- P: Weight of the painter (700N)
- A: Distance from the first rope to the painter (2 meters)
- B: Distance from the second rope to the painter (6 meters)
Now, let's calculate the moments about the point where the first rope is attached:
Clockwise moments:
Moment of the weight of the rod = W * (L/2) = 800N * (8m/2) = 3200 Nm
Moment of the weight of the painter = P * B = 700N * 6m = 4200 Nm
Anticlockwise moments:
Moment of the force in the second rope = F2 * (L - A) = F2 * (8m - 2m) = F2 * 6m
According to the principle of moments, the sum of the clockwise moments is equal to the sum of the anticlockwise moments:
3200 Nm + 4200 Nm = F2 * 6m
7400 Nm = 6F2
F2 = 7400 Nm / 6m = 1233.33 N
Therefore, the force in the second rope (F2) is approximately 1233.33 Newtons (N).
To find the force in the first rope (F1), we can subtract the force in the second rope from the weight of the system:
F1 = W - F2 = 800N - 1233.33N = -433.33 N
The negative value for F1 indicates that the force in the first rope is in the opposite direction of the weight.
Therefore, the force in the first rope (F1) is approximately -433.33 Newtons (N).