help with this physics problem please?

Question

The drawing shows three objects, with m_1 = 12.5 kg and m_2 = 27.5. They are connected by strings that pass over massless and frictionless pulleys. The objects move, and the coefficient of kinetic friction between the middle object and the surface of the table is 0.100.


link to drawing: http://i48.photobucket.com/albums/f226/buttrefly18/untitled-1.jpg

(a) What is the acceleration of the three objects?
(b) Find the tension in each of the two strings.

Answer 1

The easiest way to find the solutin is to look at FBD of each object.

Using motion as the strings moving the pulleys clock wise as positive, develop three equations:
where
T1=tension in string 1
T2 =tension in string 2
a=acceleration of all three masses since they are all connected by massless strings over massless frictionless pulleys all the accelerations are equal.

m1 FBD

T1-m1*g=m1*a

middle FBD
T2-T1-80*g*0.1=80*a

m2 FBD
m2*g-T2=m2*a

solving
T1=m1*(g+a)
T2=m2*(g-a)
T2-T1=80*(g*0.1+a)

m2*(g-a)-m1*(g+a)-80*(g*0.1+a)
=0


(m2-m1-.8)*g=(m2+m1+80)*a
a=(m2-m1-.8)*g/(m2+m1+80)
a=
(27.5-12.5-.8)*9.81/(27.5+12.5+80)

a=1.16 m/s^2 to the right

T1=12.5*(9.81+1.16)
=137 N

T2=m2*(g-a)
=27.5*(9.81-1.16)
=238 N

j

Answer 2

Draw a free body diagram of each mass. Be sure to consider the tension in the string as one of the forces (as well as gravity, and friction on the top block.) You must also remember that the tension in the string is the same for any one single string. So the tension that mass 1 experiences is the same tension that pulls on the left side of the block on top. The same is true for mass 2. Finally, remember that friction is just u*N where u is the coefficient of friction, and N is the normal (reaction) force.