Question

Figure shows a weight of 30 kg suspended at one
end of cord and a weight of 70kg applied other end
of the cord passing over a pulley. Neglecting weight
of rope and pulley, find the acceleration of the system
and tension in the cord (g=10 ms-2)​

Answer 1

Answer:

explanation:

• f=10kgm/s2
• m=w2-w1=70kg-30kg

m=40kg

• f=ma(newton's 2nd law)
• 10=40(a)
• 10/40=a
• a=0.25

Answer 2

Given: Two weights at both sides of pulley

           m₁ = 30 kg

           m₂ = 70kg

            g = 10m/s²

To Find: acceleration of the masses - a =?

              Tension - T of cord =?

Solution:

• To maintain equilibrium over the pulley, the heavier mass accelerates downwards with acceleration - "a", and the lighter mass on the other side of the pulley accelerates upwards against the gravitational force with acceleration "-a".
• The weight of the masses - mass × acceleration due to gravity - mg → force applies in the downward direction.
• The Tension applies in the upward direction in the cord above the surface of the masses towards the pulley in case of both the weights.
• Tension is the same all over the cord i.e "T".
• Therefore, two forces are acting i.e. mg (weight) and Tension.
• The sign of each force in the direction of acceleration of the mass is positive.

Equation :- addition of forces = ma

Applying the above concept:-

1. The Equation for m₁  

• mass accelerating upwards same as the direction of T, hence T - positive.

        T - m₁g = m₁a

        T - 30 × 10 = 30 × a

        T - 300 = 30a  → I

    2. The Equation for m₂

        m₂g - T = m₂a

         70 × 10 - T = 70 × a

         700 - T = 70a → II

Comparing equations I and II

   T - 300 = 30a

  - T+ 700 = 70a

          400 = 100a

          a = 400/100

          a = 4m/s²

Putting the value of a in Equation I

       T - 300 = 30a

       T = 30a + 300

       T = 30 × 4 + 300

       T = 120 + 300

       T = 420 N

Therefore acceleration of the system = 4m/s² and Tension in the cord - T = 420N.