how to find the tension in two strings, 0.75 kg meter stick is supported at the 10 cm mark and the 75 cm mark by strings. a 3.5 kg is at the 25 cm mark and a 2.5 kg is at the 90 cm mark.

Total tension on the two strings is the sum of all weights.

T1 + T2 = (0.75 + 3.5 + 2.5)*g = 66.2 N

For tension T2 (at the 75 cm mark), set total moment about the other stiring suipport point equal to zero.

T2*65 - 7.35*40 + 34.3*15 + 24.5*80 = 0

Each term is the product of a force and a measured distance from the 10 cm mark. Negative terms are clockwise moments. The 7.35 N (0.75 g) weight force is applied at the center of mass of the meter stick.

Solve for T2. Then get T1 from the sum.

To find the tension in the two strings supporting the meter stick, we can use the concept of rotational equilibrium. In order for the meter stick to be in rotational equilibrium, the sum of the torques acting on it must be zero.

Let's start by calculating the torque caused by the 3.5 kg mass at the 25 cm mark. Torque (denoted as τ) is given by the formula τ = F × r × sin(θ), where F is the force applied, r is the perpendicular distance from the pivot point, and θ is the angle between the force vector and the lever arm.

In this case, the force applied by the 3.5 kg mass is its weight, which can be calculated using the formula F = mg, where m is the mass and g is the acceleration due to gravity (approximately 9.8 m/s^2). The perpendicular distance from the pivot point to the 25 cm mark is 25 cm (or 0.25 m), and the angle between the force vector and lever arm is 90 degrees (sin(90) = 1).

Thus, the torque caused by the 3.5 kg mass is τ₁ = (3.5 kg × 9.8 m/s^2) × 0.25 m × 1 = 8.575 Nm.

Next, let's calculate the torque caused by the 2.5 kg mass at the 90 cm mark. Using the same formula, the force applied by the 2.5 kg mass is F = m × g = 2.5 kg × 9.8 m/s^2. The perpendicular distance from the pivot point to the 90 cm mark is 90 cm (or 0.9 m), and the angle between the force vector and lever arm is again 90 degrees.

Thus, the torque caused by the 2.5 kg mass is τ₂ = (2.5 kg × 9.8 m/s^2) × 0.9 m × 1 = 22.05 Nm.

To maintain rotational equilibrium, the sum of the torques acting on the meter stick must be zero. Since the torques caused by the masses are acting in opposite directions, we have:

τ₁ + τ₂ = 0

Substituting in the values we calculated:

8.575 Nm + 22.05 Nm = 0

This equation shows that the tension in the two strings must be equal and opposite to balance out the torques. Therefore, the tension in each string is equal to 8.575 N.