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A box resting on a plank

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Homework Statement


A small box of mass [itex]m_{1}[/itex] resting on a plank, of mass [itex]m_{2}[/itex] and length [itex]L[/itex], which itself rests on a frictionless, horizontal surface. Both box and board are stationary when a constant force [itex]F[/itex] is applied to the board.

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

(A picture can be found on page 26, question 13 here, though the problem is not the same)

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 [itex]F_{\text{f}}[/itex] (relative to the surface)?

Homework Equations



The equations needed to solve the problem are newtons second law (where mass = a constant), [itex] F=ma [/itex]. The inequality dealing with friction cannot be used as we are not allowed to use the coefficient of friction between the box and board.

The Attempt at a Solution



The frictional force [itex]F_{\text{f}}[/itex] must be equal to [itex]m_{1}a_{1}[/itex], where [itex]a_{1}[/itex] is the acceleration on the box. However I am looking for the acceleration relative to the table (surface).

Because the force [itex]F[/itex] accelerates the plank, and the box rests on top of it, the box's mass must be taken into account, so [itex]F=a_{2}(m_{1}+m_{2})[/itex] where [itex]a_{2}[/itex] is the acceleration on the plank. [itex]F_{\text{f}}[/itex] must act in the same direction as [itex]F[/itex] or the box would accelerate opposite to the direction of the plank and go flying off the end, so the total force on the box must be [itex]F+F_{\text{f}}=m_{1}a_{1}[/itex].

Is this reasoning correct? The question implies that [itex]F_{\text{f}}[/itex] can be expressed independently of [itex]F[/itex] or [itex]a_{2}[/itex]. Any thoughts?


Edit: I just asked the lecturer if we were supposed to assume that the box did not slip and he said "Understanding that point is part of the question!" Does that mean we are supposed to assume that, or not?
 
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Answers and Replies

  • #2
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Here are all the problems. I'll post them with my new attempts and then maybe someone can tell me if and where I am going wrong.

(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 [itex]F_{\text{f}}[/itex]?


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

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

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

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

(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 [itex]F[/itex] and [itex]F_{f}[/itex].


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

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

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

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

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

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

If [itex]F<\mu_{S}m_{1}g[/itex] then [itex]\mu_{S}m_{1}g - F > \mu_{S}m_{1}g + \mu_{S}m_{2}g [/itex]

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

[itex]F-\mu_{S}m_{1}g > \mu_{S}m_{1}g + \mu_{S}m_{2}g [/itex]

and [itex]F>\mu_{S}g(2m_{1}+m_{2})[/itex]
 
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  • #3
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For part e I may have my strict and non-strict inequalities the wrong way around...
 

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