find the acceleration and the tension in the cord.

Question

a light frictionless pulley carries a light cord to which is attached at one end a 48kg mass, the other 64kg mass. the masses are suddenly released. find the acceleration and the tension in the cord.

 

Answer 1

M = 64 kg

m = 48 kg

g = 9.8 m/s²

 

m moves up at acceleration a, while M has acceleration -a (i.e. downward).  The tensions on both masses are equal in magnitude per Newton's 3rd law.

 

T = tension

 

1st law on m:

 

(E1)   T - mg = ma

 

1st law on M:

 

(E2)   T - Mg = M(-a) = -Ma

 

Subtract equation E2 from E1 and get

 

-mg + Mg = ma + Ma

 

g(M - m) = a(m + M)

 

a = [(M - m)/(M + m)] g

 

Plug in m, M, and g, and get a in m/s².  Now plug a into either E1 or E2 and solve for T.  From E1,

 

T - mg = ma = m[(M - m)/(M + m)] g

 

T = mg + m[(M - m)/(M + m)] g

 

You can simplify, plug in the numbers, and finish from here.