Apulley is massless and frictionless. a massless inextensible string is attached to these masses: m1 = m, m2 = m, and m3 = 4 m. what is the tension t1 in the string connecting the two objects of equal mass?

T1 = 4/3 mg

Explanation:

First of all, we assume that the two objects  of equal mass, are hanging vertically, on one side of the pulley, while m3, of mass equal to 4 m, is hanging on the other side.

As the pulley is out of balance, the masses will accelerate.

As m3 is heavier, m3 wil accelerate downwards, while the set of m1 + m2 will do the same upwards.

We will take as the positive direction, the one in which acceleration is pointing, i.e., downward.

If we take as all our system to all masses, the tension T in the pulley string, is an internal force to the system, so, we can apply Newton's 2nd Law to the vertical direction, as follows:

Fnet = ma

4mg - 2mg = 6m a  ⇒ a = 1/3 g

As the acceleration is the same for all the masses, we can find the tension T, taking the 4m mass as our system:

-T + 4mg =  4m .1/3g.

Solving for T:

T = 8/3 mg

Now, if we look to the masses m1 and m2, m1, is subjected to 3 external forces, the tension T exerted by the pulley, the tension T1 exerted by m2 which is hanging from m1, and the gravity force on m1, m1g.

The sum of all these forces, according to Newton's 2nd Law, must be equal to the product of m1 times the acceleration of all the system.

Taking downward direction as positive, we can write:

-T +m₁g +T₁ = m₁ . a

Replacing by the values, and solving for T₁, we have:

-8/3mg + mg+ T₁ =-m.1/3g ⇒ T₁ = 4/3 mg

-4 on

answer: 2×10^-2kg

explanation: density= mass per volume

vol=mass/density

converting to s.i units we have;

3÷1000kg = 0.003

1.5÷100m = 0.15

0.003/0.15 = 0.02kg