P A Q

Problem:
A particle of mass 0.5kg suspended by a vertical string. A particle B of mass 0.4kg is suspended from A by means of another string.A force of 10N is applied to the upper string and the particles move upwards. Find the tension in the lower string and the acceleration of the system.

I got the tension to equal 4.44N and the acceleration to be 11.1m/s^2, it should be 1.31m/s^2. Why though?


Solution:
Particle A (mass M) is above particle B (mass m), and they are connected by the lower string.

Both strings are assumed inextensible, thus both particle A and particle B should have common kinematics.

Forces acting on particle A:
Weight (M*g): downward
Tension in upper string (T1): upward
Tension in lower string (T2): downward...why? Because of Newton's third law.

Forces acting on particle B:
Weight (m*g): downward
Tension in lower string (T2): upward
Upper string is not connected and thus doesn't act on particle B.

Newton's 2nd law equations with acceleration defined up as positive:
T1 - T2 - M*g = M*a
T2 - m*g = m*a

Two equations, two unknowns (a and T2).

Solve equation for particle B for T2:
T2 = m*(g + a)

Thus:
T1 - m*(g + a) - M*g = M*a

Solve for T1:
T1 = (M + m)*(g + a)

Solve for a:
g + a = T1/(M + m)

Thus:
a = T1/(M + m) - g

Re-substitute:
T2 = m*(g + T1/(M + m) - g)

Simplify:
T2 = T1*m/(M+m)

Summary:
T2 = T1*m/(M+m)
a = T1/(M + m) - g

Data:
T1:=10 N; m:=0.4 kg; M:=0.5 kg; g:=9.8 N/kg;

Results:
T2 = 4.444 Newtons
a = 1.311 m/s^2 

Labels: Find the acceleration of the system?

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