Find the tension in the lower portions of the string. Verify that your answers agree with part B if you set mp. Express your answer in terms of some or all of the variables m, g, and constant g.
The Correct Answer and Explanation is:
Correct Answer:
T_lower = (m2*g*(2*m1 + mp)) / (2*m1 + 2*m2 + mp)
Explanation:
This problem involves a dynamic system with two masses and a massive pulley. Based on the variable T_lower, the most probable physical setup is a mass m1 on a frictionless horizontal surface connected by a string to a hanging mass m2. The string passes over a pulley with mass mp. The term T_lower refers to the tension in the vertical portion of the string attached to the hanging mass m2.
To find this tension, we apply Newton’s second law to each component of the system. Let a be the acceleration of the masses and T_upper be the tension in the horizontal part of the string.
- For the hanging mass m2: The net force is the gravitational force minus the upward tension.
m2*g – T_lower = m2*a - For the mass m1 on the table: The net force is the tension in the horizontal part of the string.
T_upper = m1*a - For the massive pulley mp: The pulley is modeled as a solid disk with a moment of inertia I = (1/2)mp*R^2. The two different tensions create a net torque, causing it to rotate with angular acceleration α. The string does not slip, so a = αR.
The net torque is τ = (T_lower – T_upper)R.
Using τ = Iα, we get (T_lower – T_upper)R = (1/2)mp*R^2 * (a/R), which simplifies to T_lower – T_upper = (1/2)mp*a.
Now we solve this system of equations. We substitute the expressions for T_lower (from m2’s equation) and T_upper (from m1’s equation) into the pulley’s torque equation, but it is simpler to first substitute T_upper = m1*a into the torque equation to relate the tensions: T_lower – m1*a = (1/2)mp*a.
Next, we solve for acceleration by substituting this into the first equation. Rearranging the torque equation gives T_lower = m1*a + (1/2)mp*a.
Now, set the two expressions for T_lower equal:
m2*g – m2*a = m1*a + (1/2)mp*a
m2*g = a * (m1 + m2 + mp/2)
a = (m2*g) / (m1 + m2 + mp/2)
Finally, we find T_lower by substituting a back into the equation for mass m2:
T_lower = m2 * (g – a) = m2*g * (1 – m2 / (m1 + m2 + mp/2))
T_lower = m2*g * ((m1 + mp/2) / (m1 + m2 + mp/2))
Multiplying the numerator and denominator by 2 gives the final form:
T_lower = (m2*g*(2*m1 + mp)) / (2*m1 + 2*m2 + mp)
To verify, if we set mp = 0, the expression becomes T_lower = (m2*g*2*m1) / (2*m1 + 2*m2) = (m1*m2*g) / (m1 + m2), which is the correct tension for a system with a massless pulley.
