Please help me on this problem in newton's law of motion?

Question

The drawing shows three objects. They are connected by strings that pass over mass less and friction free pulleys. The object move, and the coefficient of kinetic friction between the middle object and the surface of the table is 0.100. (a) What is the acceleration of the three objects? (b) Find the tension of the in each of the two strings?

ans

(a) 0.60 m/s^2

(b) 104 N left string,

230 N right string

drawing of the problem is here

http://i68.photobucket.com/albums/i38/jester21000/help.jpg

Answer 1

Your diagram was not well scanned; the label which shows the mass of the center object is not visible.  Let's call it "m2".

m1 = 10.0 kg

m2 = ??

m3 = 25.0 kg

T1 = tension in left rope = ??

T2 = tension in right rope = ??

The secret to this problem is to apply the "F_net = ma" formula separately to each of the 3 blocks, one at a time; and notice that the "a" (the acceleration) is the same for all 3 blocks since they're tied together.

Since Block 3 is heavier than Block 1, we can make the assumption that Block 1 moves upward; Block 2 to the right; and Block 3 downward.

Block 1:

--------

F_net1 = upward force - downward force

= T1 - (m1)(g)

and since F_net1 = (m1)(a),

T1 - (m1)(g) = (m1)(a)

Block 2:

--------

F_net2 = rightward force - leftward force

= T2 - (T1 + friction)

= T2 - T1 - μ(m2)g

and since F_net2 = (m2)(a),

T2 - T1 - μ(m2)g = (m2)(a)

Block 3:

--------

F_net3 = downward force - upward force

= (m3)(g) - T2

and since F_net3 = (m3)(a),

(m3)(g) - T2 = (m3)(a)

So now you have these 3 equation to work with:

T1 - (m1)(g) = (m1)(a)

T2 - T1 - μ(m2)g = (m2)(a)

(m3)(g) - T2 = (m3)(a)

They give you m1, (hopefully) m2, m3, and μ, so that means you have 3 equations in 3 unknowns (T1, T2, and "a").  Use ordinary algebra to solve for the unknowns.