Question 8

[jgeng03]

How can the the acceleration of m2 be different from g because there is a rope between them. I think the force diagram should look be m2g = T + m1a. T = m1a and therefore m2g = 2m1a. a = (m2g)/(2m1a)

[admin]

The equations that you wrote don't work for many reasons. Instead of looking at that, let's discuss how to come up with the relevant equations.

To begin with, let's examine m2 and all the forces that it feels. m2 experiences a tension T from the cord. m2 experiences no friction and no gravity relevant to the problem*.

Thus we get:

Newton's 2nd Law - Sum F's = ma

T = m2a2

Now we examine mass m3. It experiences the tension upwards and its weight down.

Sum of F's = m3a2

m3g - T = m3a3 = 0 (taking down as +ve and following the passage which states that m3 neither rises nor falls)

Thus T = m3g

We equate both equations and get:

m3g = m2a2

a2 = (m3/m2) g


* The gravity that m2 experiences is through the y direction only, in other words, its weight m2g. Of course there will be a force normal upwards. Clearly m2 cannot accelerate in the y direction so we get:

Sum F's = m2a = 0

N - m2g = 0

N = m2g which is not relevant to this problem.

[maxsati7049]

Is there any acceleration in x direction for m3 ? giving the fact that the entire system is subjected to the same horizontal F force?

[admin]

Bump.