A uniform 100N beam is supported by two ropes. A 400N weight is suspended 1 fourth of the weight from the left. Find T1, T2 and the angle made by the left rope.
To find T1, T2, and the angle made by the left rope, we need to analyze the forces and moments acting on the beam.
Let's denote the distance from the left end of the beam to the weight as "x". Since the weight is suspended 1/4th of the way from the left, we have x = (1/4) * beam_length.
Now, let's analyze the forces. There are three forces acting on the beam:
1. T1 - the tension in the left rope.
2. T2 - the tension in the right rope.
3. Weight - the downward force due to the 400N weight.
Since the beam is in equilibrium (not accelerating), the sum of the vertical forces and the sum of the moments about any point must be zero.
1. Sum of vertical forces: T1 + T2 - Weight = 0
2. Sum of moments about the point where the left rope is attached:
The moment created by T1 is zero since the rope is attached at that point.
The moment created by T2 is (T2 * beam_length), and it acts in the clockwise direction.
The moment created by the weight is (Weight * (beam_length - x)), and it acts in the counterclockwise direction.
Therefore, the sum of the moments is (T2 * beam_length) - (Weight * (beam_length - x)) = 0
Now, let's substitute the known values:
- beam_length = total length of the beam
- Weight = 400N
- x = (1/4) * beam_length
1. Sum of vertical forces: T1 + T2 - 400 = 0
2. Sum of moments: T2 * beam_length - 400 * (beam_length - (1/4) * beam_length) = 0
Simplifying the equations, we can solve for T1, T2, and the angle made by the left rope.