Finding acceleration

We have to find the acceleration of the blocks in the above picture considering all the surfaces to be smooth and tension is in string is uniform.

Since, in the horizontal direction, there is no force, that implies that the block of mass MM does not have any acceleration.

Therefore A=0A=0.

Now, if the blocks of mass 2m2m and mm are moving with an acceleration ara_r relative to the bigger block (of mass MM) .

Now, we can write the force equations for the smaller blocks, that is 2mgT=2mar2mg-T=2ma_r and Tmg=marT-mg=ma_r.

Solving this, we get ar=g3a_r=\dfrac{g}{3}. However, the answer given to the above problem is ar=0a_r=0 as well, that is , whole system is at rest. Is my method wrong or the given answer is wrong?

Another similar problem that is bothering me is this:

Now, here similar to the previous problem I have assumed that the bigger block of mass 4m4m moves towards right and the smaller blocks move with an acceleration ara_r relative to the bigger bloc towards left.

Again in horizontal direction, net force is 00 so I wrote 4mA+m(Aar)=0    ar=5A4mA+m(A-a_r)=0 \implies a_r=5A but the given answer in this case is A=g23A=\dfrac{g}{23} and ar=6g23a_r=\dfrac{6g}{23}. Now my very first equation becomes wrong due to this. Where is the fault?

Note by Vilakshan Gupta
2 months ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Sort by:

Top Newest

Log in to reply

Acceleration is a=g3\displaystyle a = \frac{g}{3} as you claim. Maybe both blocks have equal mass mm or the answer given is wrong. :)

Aryan Sanghi - 2 months ago

Log in to reply

No, the blocks aren't of equal mass. I have one more similar problem, and again my answer does not match with the given answer. Give it a try too, I will upload it too.

Vilakshan Gupta - 2 months ago

Log in to reply

Ok, I'll give it a try. Do mention me there so that I'll get notified. :)

Aryan Sanghi - 2 months ago

Log in to reply

@Aryan Sanghi Wait, are the blocks being moved to right by a force in the question?

Aryan Sanghi - 2 months ago

Log in to reply

@Aryan Sanghi No. They are not. The system is released from rest. I have assumed the accelerations.

Vilakshan Gupta - 2 months ago

Log in to reply

@Aryan Sanghi I have uploaded it.

Vilakshan Gupta - 2 months ago

Log in to reply

@Mark Hennings Sir, please help...

Vilakshan Gupta - 2 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...