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This is a problem from my problem sheet but we haven't yet covered the material or received our books. There is a small box of mass $m_{1}$ resting on a plank, of mass $m_{2}$ and length $L$, which itself rests on a frictionless, horizontal surface. Both box and board are stationary when a constant force $F$ is applied to the board.

Take $g$ to be the acceleration due to gravity and $F_{\text{f}}$ to be the magnitude of the frictional force between the board and the box.

Assuming the coefficient of static friction between the box and the board is unknown, what is the magnitude, and the direction, of the acceleration of the box in terms of $F_{\text{f}}$ (relative to the surface)?

Instinct and experiment tells me that $F_{\text{f}}$ must act in the direction of $F$ or the box would accelerate backwards off the box, and that the magnitude of the acceleration must be $$F_{\text{f}} + F = m_{1}a_{1}$$ taking $a_{1}$ to be the acceleration of the box. So $$a_{1} = \frac{F+F_{\text{f}}}{m_{1}}$$

However, later in a later part of the question, it is implied that $F_{\text{f}}$ can be expressed independently of $F$. Is my reasoning correct, and if so how can I express it a little more convincingly. If not then why not?

Sorry for asking such a basic question but no one on my course has yet answered it.


Edit: Here are all the problems and my new answers to them so you can see what work I've done. A picture of the problem can be found on page 26, in question 13 here, though the problem is different.

(a) Assume that the coefficient of static friction between the board and the box is not known at this point. What is the magnitude, and the direction, of the acceleration of the box in terms of $F_{\text{f}}$?

If $a_{\text{box}}$ is the acceleration on the box, then $F_{\text{f}}=m_{1}a_{\text{box}}$ so $a_{\text{box}}=\frac{F_{\text{f}}}{m_{1}}$

This must act in the direction of $F$ or the box would slide or accelerate off the negative side of the board (taking the direction of $F$ to be positive)

(b) Now take the coefficient of static friction to be $\mu_{S}$. What is the largest possible magnitude of acceleration of the box? Express your answer in terms of some or all of the variables $F, \mu_{S}, m_{1}, m_{2}$ and $ g$.

So $ m_{1} a_{\text{box}} = F_{\text{f}} \leq \mu_{S} F_{n}$ where $ F_{n} $ is the normal force on the box. $ F_{n} = m_{1}g $ so $ m_{1} a_{\text{box}} \leq \mu_{S} m_{1}g$ $a_{\text{box}} \leq \mu_{S}g$ and therefore $\max{a_{\text{box}}} = \mu_{S}g$

(c) Write down the sum of all horizontal forces on the board, taking the positive direction to be towards the right. Give your answer in terms of $F$ and $F_{f}$.

The total force on the board, $F_{\text{board}}$, is equal to the accelerating force, $F$, plus the friction from the box acting on the board, $-F_{f}$ i.e. $F_{\text{board}}=F-F_{f}$

(d) Find an expression for the acceleration of the board when the force of static friction reaches its maximum possible value in terms of $F, \mu_{S}, m_{1}, m_{2}$ and $ g$.

$\max{F_{\text{f}}}=\mu_{S}m_{1}g$ so the total force on the board when static friction is at its maximum is $F_{\text{board}}=F-\mu_{S}m_{1}g$. Because the mass of the box $m_{1}$ acts through the board, $F_{\text{board}}=(m_{1}+m_{2})a_{\text{board}}$, where $a_{\text{board}}$ is the acceleration of the board.

This means $(m_{1}+m_{2})a_{\text{board}} = F-\mu_{S}m_{1}g$ so $a_{\text{board}} = \frac{F-\mu_{S}m_{1}g}{m_{1}+m_{2}}$

(e) What is the minimum value of the constant force, $F$, applied to the board, needed to ensure that the accelerations satisfy $|a_{\text{board}}|>|a_{\text{box}}|$? Express your answer in terms of some or all of the variables $\mu_{S}, m_{1}, m_{2}, g$ and $L$. Do not include $F_{\text{f}}$ in your answer.

Assuming that the force of static friction is at its maximum when slippage occurs, $\left|\frac{F-\mu_{S}m_{1}g}{m_{1}+m_{2}}\right|> |\mu_{S}g|$

If $F<\mu_{S}m_{1}g$ then $\mu_{S}m_{1}g - F > \mu_{S}m_{1}g + \mu_{S}m_{2}g $

then $-F > \mu_{S}m_{2}g$ and $ F < -\mu_{S}m_{2}g$. This implies F is negative when it is actually positive so $F \geq \mu_{S}m_{1}g$ and then

$F-\mu_{S}m_{1}g > \mu_{S}m_{1}g + \mu_{S}m_{2}g $

and $F>\mu_{S}g(2m_{1}+m_{2})$

share|improve this question
    
For part e I may have my strict and non-strict inequalities the wrong way around... –  Asa Cremin Oct 5 '13 at 23:14
    
See physics.stackexchange.com/q/79173/392 for related question. –  ja72 Oct 6 '13 at 1:07
    
so the mass of the box, $m_{1}$, doesn't need to be taken into account when calculating $F_{\text{board}}$ because we've already taken the effect of the box into account with friction? –  Asa Cremin Oct 6 '13 at 10:45

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