The system shown in the following figure is in static equilibrium and the angle

Question

The system shown in the following figure is in static equilibrium and the angle 0 = 25degree. Given that the mass is 8.80 kg and the coefficient of static friction between mass and the surface on which it rests is 0.30, what is the maximum mass that m2 can have for which the system will still remain in equilibrium? A mass mj = 3.60 kg block on a smooth tabletop is attached by a string to a hanging block of mass m2 = 2.10 kg, as shown in the figure. The blocks are released from rest and allowed to move freely. Find the magnitude of the acceleration of the blocks.

Explanation / Answer

Solving 10)

Given that,

m1 = 3.60 kg ; m2 = 2.1 kg

We need to find the acceleration of the box, let it be a.

Let T be the tension in the string.

Weight of block m2 = m2 x g = 2.1 x 9.8 = 20.58 N

T = m1 a

m2g - T = m2 a (As the block is moving freely towards down)

m2g - m1a = m2a

m2g = a(m1+m2)

a = m2g/(m1+m2) = 2.1 x 9.8 / (3.6+2.1) = 20.58/5.7 = 3.61 m/s2

Hence, the acceleration of the block = a = 3.61 m/s2

(PS: you have put multiple questions that require seperate derivation. I humbly request you to put the other question seperately. Let me know if you have any query...Thanks).

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