a 3m long light board with negligible weight is supported at each end by cables. a painter weighing 900 N stands 1m from the left cable. find the tension in each cable,
been there, done that, see above
The tensions in the cable are 300 N and 600 N
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To find the tension in each cable, we need to consider the equilibrium of forces acting on the light board.
Let's analyze the forces acting on the board:
1. The weight of the painter (900 N) acts vertically downward through the center of gravity, which is 1 meter from the left cable.
2. The tension forces in the cables act vertically upward at each end of the board.
Since the board is in equilibrium, the sum of the vertical forces must be zero. Therefore, the upward forces (tension in the cables) must balance out the downward force (weight of the painter).
Let's denote the tension in the left cable as T1 and the tension in the right cable as T2.
1. The weight of the painter (900 N) exerts a downward force at a distance of 1 meter from the left cable. Therefore, the moment (torque) caused by the painter's weight is given by:
Moment = Force x Distance
Moment = 900 N x 1 m
Moment = 900 Nm (clockwise)
2. Since the light board is in equilibrium, the total moment caused by the tension in the cables must balance out the moment caused by the painter's weight. The length of the board is 3 meters, so the moment equation is:
(T1 x 3 m) - (T2 x 0 m) = 900 Nm (clockwise)
(T1 x 3 m) = 900 Nm
T1 = 900 Nm / 3 m
T1 = 300 N
Therefore, the tension in the left cable (T1) is 300 N.
3. Since the board is also in rotational equilibrium, the total vertical force must balance out the downward force caused by the painter's weight:
T1 + T2 = 900 N
Substituting the value we found for T1:
300 N + T2 = 900 N
T2 = 900 N - 300 N
T2 = 600 N
Therefore, the tension in the right cable (T2) is 600 N.
To summarize:
- Tension in the left cable (T1) = 300 N
- Tension in the right cable (T2) = 600 N