What Determines the Acceleration of a Box on a Moving Plank?

[asaspades]

Homework Statement

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.

(A picture can be found on http://www.columbia.edu/~crg2133/Files/Physics/8_01/lotsOfProbs.pdf, 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 F_{\text{f}} (relative to the surface)?

Homework Equations

The equations needed to solve the problem are Newtons second law (where mass = a constant), F=ma. 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 F_{\text{f}} must be equal to m_{1}a_{1}, where a_{1} is the acceleration on the box. However I am looking for the acceleration relative to the table (surface).

Because the force F accelerates the plank, and the box rests on top of it, the box's mass must be taken into account, so F=a_{2}(m_{1}+m_{2}) where a_{2} is the acceleration on the plank. F_{\text{f}} must act in the same direction as F 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 F+F_{\text{f}}=m_{1}a_{1}.

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

 

[asaspades]

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 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})

 

[asaspades]

For part e I may have my strict and non-strict inequalities the wrong way around...