Laws of Motion - Problem

  • #1
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Homework Statement


Two cardboard boxes full of books are in contact with each other on a friction less table.Box H has twice the mass of Box G. If you push on box G with a horizontal force F, then box H will experience a net force of?



2. The attempt at a solution

Let book G's mass be = m
Then book H's mass is = 2m

Let the force applied on book g (push) be F.

Then,
F = ma -| From second law of motion

Now since the block will start moving with a velocity of at (Where T is time).



Now using the first equation of motion: v = u + at, (where u = 0)
at = c(t)

Where C is a variable for acceleration of the body.

This implies as

a = c.
Thus the body G has a mass of m and acceleration of c = a.

Hence the force applied on Book H

ma = 2ma2


Where a2 is the acceleration attained by book H


The answer given is 2/3 F please explain all steps clearly,
Thanks


 

Answers and Replies

  • #2
Orodruin
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Then, F = ma -| From second law of motion
No, this is wrong. The force you are pushing with is not the only force on box G. The force appearing in Newton's second law is the total force on the system.

Think about how the force F must relate to the acceleration of the entire system.

please explain all steps clearly,
Nobody here will give you a full solution. We will help you with hints and pointing out the places where you go wrong, but you need to do the work yourself. Also see the homework guidelines.
 
  • #3
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So then,
$$ \rm {F = F_s} $$

$$ F_s = (m + 2m)a = 3ma \\ F_h = 2ma \implies F_h = 2 \frac{F}{3}$$

Is that the way then?

Nobody here will give you a full solution. We will help you with hints and pointing out the places where you go wrong, but you need to do the work yourself. Also see the homework guidelines.
I respect that.

However would you clear one doubt for me? Why do we consider the whole system and not consider $$ \rm{Body G}$$ individually? My idea was that $$ F $$ would affect $$\rm{Body G}$$ which would then push $$\rm{Body H}$$
 
  • #4
TSny
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So then,
$$ \rm {F = F_s} $$

$$ F_s = (m + 2m)a = 3ma \\ F_h = 2ma \implies F_h = 2 \frac{F}{3}$$

Is that the way then?
Looks good to me.
However would you clear one doubt for me? Why do we consider the whole system and not consider $$ \rm{Body G}$$ individually?
You can approach it this way and it makes a good exercise. If you're going to treat each body separately, make sure you first draw a good free body diagram for each body separately.
 

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