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The cart is rolls frictionless on the table. It has a mass of $1 kg$. Attached to it are 2 strings, that go through two frictionless sheaves. The weights have masses as in the picture.

a) How large is the acceleration of the cart?

b) How large is the tension in the strings?

c) We replace the cart with a box that also has a mass of $1 kg$, it then accelerates with $1.5m/s^2$, find the friction force between the table and the box."

I tried doing this one, and got every single answer wrong, even though I thought I knew how to do it. This is how I did it:

a) The force pulling the cart to the left is equal to the force pulling the weights down since the cart is attached with a string through a sheave. The force pulling left is equal to: $F = 1 kg \times 10 m/s^2 = 10N$ and the force pulling right is $F = 2kg m/s^2 = 20N$, therefore the resulting force is $10N$ to the right. Since the cart has a mass of $1kg$ it accelerates with $10m/s^2$.

b) The tension on the left string is $10N$ (from above) and on the right one is $20N$ (also from above).

c) This one I just have no idea how to do.

The answers given in the book are: a) $2.5m/s^2$, b) $12.5N$ and $15N$, c) $4N$

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1 Answer 1

up vote 2 down vote accepted

No, your solution is wrong. You found the resultant force and took the cart's mass into consideration to calculate the acceleration, but it is wrong. You have to take the mass of whole sytem on which the resultant force is acting, so in this case resultant force$f = 10N$, so the acceleration will be $\frac{f}{m_1+m_2+m_3}$, where $m_1, m_2,$ and $m_3$ are the mass of the two blocks and the mass of the cart. So the solution will be $\frac{10}{4kg} = 2.5m/s^2$ which is answer to your question "a". Now moving to b, look at the image:enter image description here

So as you can see there are 4 forces acting as $m_3g$ will have no effect, as there is no friction in this case so there will be different tension in the string at two different side. So $T_1$ will be $12.5N$ and $T_2$ will be $15N$. Now moving to question c. The frictional force will be $1kg*g*\mu$ where $\mu$ is the frictional coefficient. Now it is given that acceleration of the block is $1.5m/s^2$ so the equation will be $a = \frac{2g - g - f}{2kg + 1kg + 1kg}$ where $f$ is the frictional force. So substituting the values we get $1.5m/s^2 = \frac{10 - f}{4}$ so $ 6 = 10 - f $ so $f = 4N$. So I think it would clear your problem.

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Thank's Zetta Suro for your attention –  Akash Mar 20 '13 at 15:05

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