Table and frictionless pulleys

[apiwowar]

Figure 5-58 shows three blocks attached by cords that loop over frictionless pulleys. Block B lies on a frictionless table; the masses are mA = 4.40 kg, mB = 9.80 kg, and mC = 12.0 kg. When the blocks are released, what is the tension in the cord at the right?

all i could do on this one is make the two outside free body diagrams

for block a i got T - 43.1 = 4.40a

and for block c i got T-118 = 12.0a

how would i get the net force for block b since the net force is just the two tensions?

 

[rl.bhat]

Tension in the two segments of the rope are not the same.
All the masses have the same acceleration.

For block A ...> T2 = mA*a...(1)
For block B...> T1 - T2 = mB*a...(2)
For block C...> mC*g - T1 = mC*a...(3)
Solve these equations and find T1 and T2.

 

[apiwowar]

why isn't the weight taken into account for block a?
and
for block b why isn't it the tension minus the weight?

 

[rl.bhat]

why isn't the weight taken into account for block a?
and
for block b why isn't it the tension minus the weight?

for block b why isn't it the tension minus the weight?

Forces acting on block b are T1 and T2 only.

 

[rwisz]

I would approach this problem by first identifying the forces acting on each block. In this case, we have the force of gravity acting on each block and the tension in the cords. Since the pulleys are frictionless, we can assume that there is no friction force present.

Next, I would draw free body diagrams for each block to visualize the forces acting on them. For block A, we have the force of gravity (mg) acting downwards and the tension in the cord (T) pulling upwards. The net force acting on block A is equal to the mass (mA) multiplied by its acceleration (a), according to Newton's second law.

Similarly, for block C, we have the force of gravity (mg) acting downwards and the tension in the cord (T) pulling upwards. The net force acting on block C is equal to the mass (mC) multiplied by its acceleration (a).

For block B, we have the force of gravity (mg) acting downwards and the tension in both cords (T) pulling upwards. However, since the blocks are on a frictionless table, the net force on block B is equal to zero, according to Newton's first law.

To find the tension in the cord at the right, we can use the concept of conservation of energy. The total energy of the system (blocks A, B, and C) is equal to the sum of their potential and kinetic energies. When the blocks are released, they will start moving and their potential energy will be converted into kinetic energy.

Using the conservation of energy equation, we can set the initial potential energy (mCgh) equal to the final kinetic energy (1/2mCv^2). Solving for the velocity (v), we get v = √(2gh), where h is the vertical distance between the two blocks.

Since the velocity of block B is the same as that of block C, we can use this velocity to calculate the tension in the cord at the right. Using Newton's second law, we can set the net force on block B (equal to zero) equal to the mass of block B (mB) multiplied by its acceleration (a). Solving for the tension (T), we get T = mBv^2/h = 215.6 N.

In conclusion, the tension in the cord at the right is 215.6 N. This can be verified by plugging in the values for