The system starts at rest. A mass of 20kg (m1) is connected to a pulley of uniform mass m=5kg and radius 10cm and a mass of 30kg (m2) hangs on the other side of the pulley hanging 2m above the ledge

Find the tension of the strings
I tried to find the acceleration by finding the sum of the forces on both blocks and the sum of the torque:
F1 = T1 - m1*g = m1*a
so T1 = m1a + m1g
F2 = m2*g - T2 = m2*a
so T2 = m2*a - m2*g
sum torque = r*T2 - r*T1 = I*A^2 [because T2>T1? and A is angular acceleration]
I = 1/2*m*r^2 and A = a/r
all the values of r cancel,
T2 - T1 = 1/2*m*a
m2*a - m2*g - m1*a - m1*g = 1.2*m*a
solving for a gives
a = (m2*g - m1*g)/(1/2*m + m2 - m1)
I got 7.84 m/s^2 with the given values, and substituting into equations for T1 and T2
T1 = m1*a + m1*g
= 20*7.84 + 20*9.8
= 353N
T2 = m2*g - m2*a
= 30*9.8 - 30*7.84
= 58.8
But this makes no sense because T2 should be greater than T1? Can you please tell me where i went wrong

It seems like there was an error in your calculation. Let's go through the steps again to find the correct tension in the strings.

To find the tension in the strings, we need to consider the forces acting on each block. Let's assume that the positive direction is upwards.

For Block 1 (m1):
The forces acting on Block 1 are the tension in the string (T1) and the weight of m1 (m1 * g). The net force acting on Block 1 is given by:

F1 = T1 - m1 * g

For Block 2 (m2):
The forces acting on Block 2 are the weight of m2 (m2 * g) and the tension in the string (T2). The net force acting on Block 2 is given by:

F2 = m2 * g - T2

Since the pulley is massless and frictionless, we can assume that the tension in both strings is the same (T1 = T2 = T).

Using Newton's second law (F = ma) for each block, we can write two equations:

m1 * a = T - m1 * g
m2 * g - T = m2 * a

To eliminate the tension (T) and solve for acceleration (a), we need to solve these two equations simultaneously.

Rearranging the equations, we get:

T = m1 * (g + a)
T = m2 * (g - a)

Equating the two expressions for T, we have:

m1 * (g + a) = m2 * (g - a)

Expanding and rearranging, we get:

a * (m2 + m1) = g * (m2 - m1)

Now we can solve for the acceleration (a):

a = g * (m2 - m1) / (m2 + m1)

Substituting the given values, we have:

a = 9.8 * (30 - 20) / (30 + 20)
a = 9.8 * 10 / 50
a = 1.96 m/s^2

Now we can substitute this value of acceleration into the equations for tension (T1 and T2) to find the tension in the strings:

T1 = m1 * (g + a)
T1 = 20 * (9.8 + 1.96)
T1 = 20 * 11.76
T1 = 235.2 N

T2 = m2 * (g - a)
T2 = 30 * (9.8 - 1.96)
T2 = 30 * 7.84
T2 = 235.2 N

Thus, the correct values for the tensions in the strings are T1 = T2 = 235.2 N.

To calculate the tension in the strings, you correctly found the acceleration of the system using Newton's second law for each block. However, there is a mistake in your calculation of the tensions.

Let's go through the steps again and find the correct values for T1 and T2.

1. Find the acceleration (a) of the system:
Since the system starts at rest, the total net force on the system is equal to the total mass times acceleration.
Sum of forces on m1 (F1):
F1 = T1 - m1 * g
Sum of forces on m2 (F2):
F2 = m2 * g - T2

Setting F1 = F2 and solving for a:
T1 - m1 * g = m2 * g - T2
T1 + T2 = (m2 + m1) * g

Sum of torques (I * α):
Sum of torques = (r * T2) - (r * T1) = (0.5 * m * r^2) * α
Since the pulley has a radius of 10 cm = 0.1 m:
0.1 * (T2 - T1) = (0.5 * 5 * 0.1^2) * (a / 0.1)
0.1 * (T2 - T1) = 0.05 * a

2. Substitute the acceleration into the equation:
T1 + T2 = (m2 + m1) * g
T1 + T2 = (30 kg + 20 kg) * 9.8 m/s^2
T1 + T2 = 490 N

0.1 * (T2 - T1) = 0.05 * a
0.1 * (T2 - T1) = 0.05 * 7.84 m/s^2
0.1 * (T2 - T1) = 0.392 N

3. Solve the system of equations:
T1 + T2 = 490 N
0.1 * (T2 - T1) = 0.392 N

Solving these equations simultaneously will give us the correct values for T1 and T2.
Let's solve these equations:

Expand the second equation:
0.1 * T2 - 0.1 * T1 = 0.392

Rearrange the first equation:
T1 = 490 - T2

Substitute T1 in the expanded equation:
0.1 * T2 - 0.1 * (490 - T2) = 0.392

Simplify and solve for T2:
0.1 * T2 - 0.1 * 490 + 0.1 * T2 = 0.392
0.2 * T2 - 49 = 0.392
0.2 * T2 = 49.392
T2 = 49.392 / 0.2
T2 = 246.96 N

Substitute T2 into the first equation to solve for T1:
T1 = 490 - T2
T1 = 490 - 246.96
T1 = 243.04 N

Therefore, the correct tension in the first string (T1) is 243.04 N, and the correct tension in the second string (T2) is 246.96 N.