A pulley is massless and frictionless. 3 kg, 2 kg (on the left), and 6 kg (on the right) masses are suspended What is the tension T1 in the string be-

tween the two blocks on the left-hand side
of the pulley?

Can I solve this like if it was two masses?

My prof gave an example T=m1g=m1a
m2g-t+m2a
then we solved for T in equation one and plugged it into equation two... in that problem the Tension was the same on both sides but in the picture on the problem they have labeled them T1, T2 and T3 so I'm not sure

T2 and T3 (around the pulley) should be equal. You can use this to solve for the tension=T2=T3 just like the prof's example.

After that, use the same principle to solve for T1 between masses m1 and m2.
Note that m2 is pulled with a tension of T=T2=T3.

Well, solving this problem will definitely require some mental gymnastics. But hey, who needs a real gym when you can exercise your brain, right?

Now, let's address your doubt. Can you solve this like if it was two masses? Well, you can certainly try, but remember, this is not your ordinary problem. It's a circus-themed problem with clown physics involved.

In all seriousness, since the pulley is massless and frictionless, we can assume that the tensions on both sides of the pulley are equal. So, T1 = T2 = T3. This makes sense because you don't want one side of the pulley feeling left out, right?

Now, using your professor's example, let's play along. Start by applying Newton's second law to the 3 kg mass on the left. We have T1 - (3 kg)(9.8 m/s^2) = (3 kg)(a).

Next, let's move to the 2 kg mass on the left. We have the tension T1 going downwards and gravitational force going upwards. So, T1 - (2 kg)(9.8 m/s^2) = -(2 kg)(a).

Last but not least, let's take a look at the 6 kg mass on the right. We have the tension T3 going upwards and the gravitational force going downwards. So, T3 - (6 kg)(9.8 m/s^2) = (6 kg)(a).

Now, stick with me because here comes the fun part. Since we know T1 = T2 = T3, we can say T1 = T2 = T3 = T (I know, I know, this is getting repetitive).

Now, you got yourself three shiny equations to solve for the acceleration. Plug this newfound knowledge into the three equations you already have, solve them simultaneously, and voila! You'll find the value of T1.

Just remember, laughter is the best way to cope with confusing physics problems. So, embrace your inner clown and enjoy the circus of physics. Good luck!

Yes, you can solve this problem using a similar approach as you described.

First, let's analyze the forces acting on the 3 kg and 2 kg masses. The 3 kg mass is pulled downwards by its weight (mg), and there is tension T1 in the string acting upwards. Similarly, the 2 kg mass is pulled downwards by its weight (2g) and there is tension T1 acting upwards.

Since the pulley is massless and frictionless, the tensions on both sides of the pulley must be equal. Therefore, T1 is the tension in the string between the two blocks on the left-hand side of the pulley.

To find the value of T1, we can write the equation of motion for the 3 kg mass:

T1 - 3kg * g = 3kg * a
(Tension upwards - Weight downwards = mass * acceleration)

Similarly, we can write the equation of motion for the 2 kg mass:

2kg * g - T1 = 2kg * a

Now we have two equations with two unknowns (T1 and a). We can solve these equations simultaneously to find the value of T1.

Yes, you can definitely solve this problem using a similar approach as you mentioned.

First, let's consider the forces acting on each of the masses. The 3 kg mass is subject to its weight (mg) downward, and tension T1 acting upward. The 2 kg mass is also subject to its weight (mg) downward and tension T2 acting upward. Finally, the 6 kg mass is subject to its weight (mg) downward and tension T3 acting upward.

Since the pulley is assumed to be massless and frictionless, the tensions on both sides of the pulley (T1 and T2) are equal in magnitude. Similarly, the tension T3 is equal in magnitude to the net force on the right side of the pulley.

Using Newton's second law, we can write the following equations:

For the 3 kg mass:
m1 * g - T1 = m1 * a1

For the 2 kg mass:
m2 * g - T2 = m2 * a2

For the 6 kg mass:
T3 - m3 * g = m3 * a3

Since the system is connected, the acceleration of each mass will have the same magnitude and direction. Thus, a1 = a2 = a3 = a.

Now, let's solve these equations step by step:

1. Solve for T1:
From the first equation, we have:
T1 = m1 * g - m1 * a

2. Solve for T2:
Since T1 = T2, we substitute the value of T1 into the second equation:
m2 * g - T2 = m2 * a
T2 = m2 * g - m2 * a

3. Solve for T3:
From the third equation, we have:
T3 = m3 * g + m3 * a

Now, you have expressions for T1, T2, and T3. You can substitute the given masses (m1, m2, m3) and acceleration (a) into these equations to find the values of T1, T2, and T3.

It's important to note that in this case, the value of T1 is the tension in the string between the two blocks on the left-hand side of the pulley. The values of T2 and T3 represent the tensions on the right-hand side of the pulley.