physics tension problem?

Question

The suspended 2:1 kg mass on the right is
moving up, the 1:7 kg mass slides down the
ramp (23 degree angle), and the suspended 8:7 kg mass on the
left is moving down. The coeficient of friction
between the block and the ramp is 0:14meu
The acceleration of gravity is 9:8 m/s2 : The
pulleys are massless and frictionless.

What is the acceleration of the three block
system?
ANSWER: acceleration is 5.52340
so how do tension?
What is the tension in the cord connected to the 2:1 kg block? What is the tension in the cord connected to the 8:7 kg block? Answer in units of N.

Answer 1

Let the tension of string attached to 8.7 kg be T1 and that attached to 2.1 be T2. Let common acceleration be a. Cos 23 = 0.921

equations are:

8.7a = 8.7g - T1
1.7a = T1 - T2 - 1.7gx0.14xCos 23 = T1 - T2 - 0.219g
2.1 a= T2 - 2.1 g

Solve thees equations first for a then for T1 and T2

I f you have doubt about the validity of any one of the above equations you may ask your specific doubts.

Answer 2

I don't like numbers, so I will replace the package weight with W, and the given angle with theta. Tension in upper cable is T1 Tension in lower cable is T2 The system is not accelerating, so all forces must add up to zero. Vertical force balance: T1*sin(theta) = W Horizontal force balance: T1*cos(theta) = T2 Thus: T1 = W/sin(theta) T2 = T1*cos(theta)/sin(theta) = W/tan(theta) Data: W:=500 lbs; theta:=30 deg; Results: T1 = 1000 lbs T2 = 866 lbs

Answer 3

I think “Let’slear” guy has not taken into account the contribution of the slab on the ramp, i.e.
1.7a = T1 - T2 - 1.7gx0.14xCos 23 +1.7gxSin 23