A uniform plank of length L and mass M is supported by two vertical ropes, one at each end. A painter of mass m stands at a distance x from the left end of the plank. What is the tension in the left rope?
Correct: A
This is a static equilibrium problem requiring both translational and rotational equilibrium conditions.
Let T_left be the tension in the left rope and T_right be the tension in the right rope.
Forces acting on the plank:
1. Weight of the plank (Mg) acting downwards at its center (L/2 from either end).
2. Weight of the painter (mg) acting downwards at distance x from the left end.
3. Tension T_left acting upwards at the left end (x=0).
4. Tension T_right acting upwards at the right end (x=L).
**1. Translational Equilibrium (ΣFy = 0):**
T_left + T_right - Mg - mg = 0
T_left + T_right = (M + m)g (Equation 1)
**2. Rotational Equilibrium (Στ = 0):**
To find T_left, it's convenient to choose the right end (x=L) as the pivot point to eliminate T_right from the torque equation.
Torques about the right end (x=L):
- Torque due to T_left: T_left * L (counter-clockwise, positive)
- Torque due to Mg: Mg * (L - L/2) = Mg * (L/2) (clockwise, negative)
- Torque due to mg: mg * (L - x) (clockwise, negative)
Στ_right = 0
T_left * L - Mg * (L/2) - mg * (L - x) = 0
T_left * L = Mg * (L/2) + mg * (L - x)
Solve for T_left:
T_left = (Mg * L/2 + mg * (L - x)) / L
T_left = (Mg * L/2) / L + (mg * (L - x)) / L
T_left = Mg/2 + mg * (1 - x/L)
T_left = ( (M/2) + m(1 - x/L) )g
The correct choice is A.