2-accelerating bodies

[plzNOCarribbean]

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Hey everyone, just had a question regarding two accelerating bodies and how to solve problems of this nature. The concept is addressed at the end of chapter 2 (mechanics I) in the TPR PS book.

It says that whenever you are given problems with 2 accelerating bodies, you should have 2 separate free body diagrams, which results in 2 equations and 2 unknowns (in the example below, it should be T, tension, and a, acceleration.

http://www.physics247.com/physics-homework-help/miforcer.jpg

The book says to add up the two equations, so that 1 variable cancels out, (which would be the tension). In the example above, lets say the large mass that is dangling from the pulley is M, which is 40 kg. The mass on the ramp is m, which weighs 20 kg. In this system, the Tension in the rope associated with M is T1, and the tension with the mass on the ramp is T2. Theta=30 degrees.

Given this, Fnet1 = W-T1 and Fnet 2=T-mgsintheta
well cant we just omit the two tensions if we assign positive (+) as the direction of the motion of the objects. So, the T1 would be negative tension (since its in the opposite direction of the motion), and T2 would be positive Tension (since its in the direction of the motion of the object)

That way, when we add the two equations together, we simply have

Fnet1 + Fnet2 = W-mgsin theta
a (m+M) = W-mgsintheta
a = w-mgsintheta/ (m+M)

a= 400 N-100 N/ (60 kg)
a= 5 m/s^2

Is this the trick to solving these problems? can you just cancel out the tensions like I did in this example, since when your assigning positive & negative directions with respect to EACH INDIVIDUAL OBJECT on the FBD, the two tensions cancel out? Thanks! 😀

 

[Blaughable]

Yep that's pretty much the idea. What I do is first decide which way is positive direction and which way is negative direction. I chose going up the ramp, over the pulley and down the string as positive (pulleys just curve your axes).

Then I do a free body diagram for each block and sum their forces in the direction of concern:

sum(F1) = Ma = Mg - T
sum(F2) = ma = T - mgsintheta

Because Tension is constant throughout the string, next solve for T:

Mg - Ma = T
ma + mgsintheta = T

and set them equal to each other thus eliminating the unknown tension:

Mg - Ma = ma + mgsintheta

Solve for acceleration:

ma + Ma = Mg - mgsintheta
a(m + M) = Mg - mgsintheta
a = (Mg - mgsintheta) / (m + M)