two masses m1=400.g and m2 =600.g are connected with a light string which goes over a frictionless pulley of negligible mass. the two systems are released from rest. a) what is the acceleration of each mass? b) what is the tension force in the string? and c) what is the support force in the pivot of the pulley?

for part c, the support force is double the tension. The tension force is what "pushes" down on the pulley. So, it will be 2T.

To find the acceleration of each mass, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the product of its mass and acceleration.

Let's assume that the direction to which m1 moves is positive and m2 moves downward. Also, let's denote the acceleration of m1 as a1 and the acceleration of m2 as a2.

a) Acceleration of each mass:
For m1: The only force acting on m1 is the tension force in the string, directed downward. Therefore, the net force on m1 is equal to the tension force. Applying Newton's second law to m1:
m1 * a1 = T

For m2: The gravitational force acts downward on m2, and the tension force in the string acts upward. The net force on m2 is equal to the difference between these two forces. Applying Newton's second law to m2:
m2 * a2 = m2 * g - T

Since the two masses are connected by a light string, the acceleration of both masses will be the same. Therefore, a1 = a2 = a.

b) Tension force:
We can solve the above two equations simultaneously to find the tension force. Substitute T from the first equation into the second equation:
m2 * a = m2 * g - m1 * a
m2 * a + m1 * a = m2 * g
a * (m2 + m1) = m2 * g
a = (m2 * g) / (m2 + m1)

Now, substitute this value of a back into the first equation to find the tension force:
T = m1 * a
T = m1 * [(m2 * g) / (m2 + m1)]

c) Support force on the pulley:
The support force in the pivot of the pulley is equal to the total force required to balance the masses.

For m1: The support force is directed upward and cancels out the tension force.

For m2: The support force is directed downward and is equal to the sum of the gravitational force and the tension force.

Therefore, the support force on the pulley is given by:
Support force = T + m2 * g

Now, you can substitute the given values of the masses m1 = 400 g = 0.4 kg and m2 = 600 g = 0.6 kg and the acceleration due to gravity g = 9.8 m/s^2 into the equations to get the numerical values for the acceleration, tension force, and support force.

To find the answers to these questions, we can use Newton's second law of motion, which states that the net force acting on an object is equal to its mass multiplied by its acceleration.

a) Acceleration of each mass:
Since the two masses are connected by a light string over a frictionless pulley, they will move together. The force due to gravity acts on each mass, and since there is tension in the string, there will be a net force on both masses.

For m1 (400 g or 0.4 kg):
The net force acting on m1 is the force of gravity (mg) minus the tension force (T). So, we have:
Net force on m1 = mg - T

For m2 (600 g or 0.6 kg):
The net force acting on m2 is the force of gravity (mg) plus the tension force (T). So, we have:
Net force on m2 = mg + T

Since the two masses move together, their accelerations will be the same. Let's call it "a".

For m1:
ma = mg - T

For m2:
ma = mg + T

Adding these two equations together, we get:
2ma = 2mg
a = g

Therefore, the acceleration of each mass is equal to the acceleration due to gravity (g), which is approximately 9.8 m/s^2.

b) Tension force in the string:
To find the tension force in the string, we can use one of the equations we derived earlier. Let's use the equation for m1:
ma = mg - T

Substituting the values we know:
(0.4 kg)(9.8 m/s^2) = (0.4 kg)(9.8 m/s^2) - T
T = (0.4 kg)(9.8 m/s^2) - (0.4 kg)(9.8 m/s^2)
T = (0.4 kg)(9.8 m/s^2)
T ≈ 3.92 N

Therefore, the tension force in the string is approximately 3.92 Newtons.

c) Support force at the pivot of the pulley:
Since the pulley is frictionless, there is no torque acting on it. The sum of the torques on the pulley must be zero.

The torque due to the support force at the pivot is equal and opposite to the torque due to the tension force in the string.

The torque due to the support force (τs) is given by:
τs = r * Support force (where r is the radius of the pulley)

The torque due to the tension force (τt) is given by:
τt = r * Tension force

Since the two torques are equal in magnitude but opposite in direction, we have:
τs = -τt

Therefore, the support force at the pivot of the pulley is equal in magnitude but opposite in direction to the tension force in the string, which is approximately 3.92 N.
Hence, the support force at the pivot is also approximately 3.92 N.

draw the diagram, put the large mass on the right. Now write a force equation near the pulley, clockwise direction positive.

m2*g-m1*g=totalmass*a=(m1+m2)*a
solve for acceleration, that is the accelerlation of each mass in the clockwise direction (one goes up, one goes down in vertical reference)

b. tension. at the top of m2.
downward force m2*g
retarding force= m1*g

F=ma
m2*g-m2*a = tension

at the pulley, one has m1+ m2 handing from it, force=(M1+M2)*g