Three masses are connected by a light string that passes over a frictionless pulley. On the right is tension one holding a 6.5 kg mass. On the left, the top mass of 4.05 kg is held by tension 2 and below that mass is tension 3 holding a 2.4kg mass.
(a) What is the magnitude of the acceleration of the system?
(b) What are the magnitudes of the three tensions in the string?
To find the magnitude of the acceleration of the system, we need to use Newton's second law and consider the forces acting on each mass.
Let's define the positive direction as upwards. The forces acting on the 6.5 kg mass are the force of gravity (mg) acting downwards and tension 1 (T1) acting upwards. Therefore, the net force on the 6.5 kg mass is given by:
Net force on 6.5 kg mass = T1 - mg
Now, considering the 4.05 kg mass, there are three forces acting on it. The force of gravity (mg) is acting downwards, tension 2 (T2) is acting downwards, and tension 1 (T1) is acting upwards. Therefore, the net force on the 4.05 kg mass is given by:
Net force on 4.05 kg mass = T2 - T1 - mg
Finally, considering the 2.4 kg mass, there are two forces acting on it. The force of gravity (mg) is acting downwards, and tension 3 (T3) is acting upwards. Therefore, the net force on the 2.4 kg mass is given by:
Net force on 2.4 kg mass = T3 - mg
Since the masses are connected by a light string, the tensions in the string are the same throughout. Therefore, we can conclude that T1 = T2 = T3 = T.
Now, we can write the equations of motion for each mass:
For the 6.5 kg mass:
T1 - (6.5 kg)(9.8 m/s^2) = (6.5 kg)(a)
For the 4.05 kg mass:
T2 - T1 - (4.05 kg)(9.8 m/s^2) = (4.05 kg)(a)
For the 2.4 kg mass:
T3 - (2.4 kg)(9.8 m/s^2) = (2.4 kg)(a)
To solve these equations, we can substitute T1 = T2 = T3 = T and solve for T and a.
To find the magnitude of the acceleration of the system, we can use Newton's second law of motion, which states that the net force on an object is equal to its mass multiplied by its acceleration, F = ma.
(a) To solve for the acceleration, we need to consider the forces acting on each mass separately. Assuming downward motion is positive, we have:
For the 6.5 kg mass:
Tension 1 - weight = mass * acceleration
For the 4.05 kg mass:
weight - Tension 2 = mass * acceleration
For the 2.4 kg mass:
weight - Tension 3 = mass * acceleration
The weight of an object is given by weight = mass * acceleration due to gravity (g).
Now, let's find the weight of each mass:
Weight of the 6.5 kg mass = 6.5 kg * 9.8 m/s^2 = 63.7 N
Weight of the 4.05 kg mass = 4.05 kg * 9.8 m/s^2 = 39.69 N
Weight of the 2.4 kg mass = 2.4 kg * 9.8 m/s^2 = 23.52 N
Now we can rewrite the equations for each mass:
For the 6.5 kg mass: Tension 1 - 63.7 N = 6.5 kg * acceleration
For the 4.05 kg mass: 39.69 N - Tension 2 = 4.05 kg * acceleration
For the 2.4 kg mass: 23.52 N - Tension 3 = 2.4 kg * acceleration
Next, we can sum up the forces and set them equal to the net force, which is the sum of all the forces acting on the system.
Sum of forces = Tension 1 - Tension 2 - Tension 3
Since the force due to tension acts in opposite directions for Tension 1 and Tension 2, their sum is equal to Tension 2 - Tension 1.
Sum of forces = (Tension 2 - Tension 1) - Tension 3
Now we can set this equal to the net force:
Sum of forces = Tension 2 - Tension 1 - Tension 3 = mass * acceleration
Substituting the masses and combining terms, we get:
(39.69 N - Tension 2) - Tension 1 - Tension 3 = (6.5 kg + 4.05 kg + 2.4 kg) * acceleration
39.69 N - Tension 2 - Tension 1 - Tension 3 = 12.95 kg * acceleration
Simplifying the equation, we have:
39.69 N - Tension 2 - Tension 1 - Tension 3 = 12.95 kg * acceleration
Now, we can solve for acceleration using this equation.
(b) To find the magnitudes of the tensions in the string, we need to solve the system of equations formed by the forces acting on each mass. The equations we obtained earlier were:
For the 6.5 kg mass: Tension 1 - 63.7 N = 6.5 kg * acceleration
For the 4.05 kg mass: 39.69 N - Tension 2 = 4.05 kg * acceleration
For the 2.4 kg mass: 23.52 N - Tension 3 = 2.4 kg * acceleration
By solving these equations simultaneously, we can find the values of Tension 1, Tension 2, and Tension 3.