Page History

{composition-setup}{composition-setup}
{table:border=1|frame=void|rules=cols|cellpadding=8|cellspacing=0}
{tr:valign=top}
{td:width=350|bgcolor=#F2F2F2}
{live-template:Left Column}
{td}
{td}


!pushingboxpushing2boxes.png!

{deck:id=bigdeck}
{card:label=Part A}

h3. Part A

{excerpt}A person pushes a box of mass 15 kg along a smooth floor by applying a perfectly horizontal force _F_.  In the process, the 15 kg box pushes against another box with a mass of 10 kg and causes it to move.{excerpt}  The boxboxes acceleratesaccelerate horizontally at a rate of 2.0 m/s{color:black}^2^{color}.  

{card:label=Part A}

h3. Part A

What is the magnitude of _F_?

h4. Solution

{toggle-cloak:id=sysa} *System:* {cloak:id=sysa} Both boxes Boxtogether as a single [point particle].
!pushing2boxessys1.png!
{cloak}

{toggle-cloak:id=inta} *Interactions:* {cloak:id=inta}External influences from the person (applied force) the earth (gravity) and the floor (normal force).{cloak}

{toggle-cloak:id=moda} *Model:* {cloak:id=moda}[Point Particle Dynamics].{cloak}

{toggle-cloak:id=appa} *Approach:*  

{cloak:id=appa}

The word *smooth* in the problem statement is a keyword, telling us that the floor exerts no horizontal force on the box.  Thus, the only external _horizontal_ force acting on the system composed of the two boxes together is that applied by the person. This implies that[Newton's 2nd Law|Newton's Second Law] for the _x_ direction can be written:

{latex}\begin{large}\[ \sum F_{x} = F = mam_{\rm system}a_{x} = (\mbox{15 kg} + \mbox{10 kg})(\mbox{2 m/s}^{2}) = \mbox{3050 N}\] \end{large}{latex}

{cloak}

{card}
{card:label=Part B}

h3. Part B

A person pushes a box of mass 15 kg along a smooth floor by applying a perfectly horizontal force _F_.  The box moves horizontally at a constant speed of 2.0 m/s in the direction of the person's applied force.  What is the magnitude of _F_?

h4. SolutionWhat is the force applied on the front (10 kg) box by the rear (15 kg) box?

h4. Solution

{toggle-cloak:id=sysb} *System:*  {cloak:id=sysb}This time, we will focus only on the front box.
!pushing2boxessys2.png!
{cloak}

{toggle-cloak:id=intb} *Interactions:* {cloak:id=intb}The front box is subject to external influences from the rear box, the earth (gravity) and the floor (normal force). {cloak}

{toggle-cloak:id=sysbmodb} *System, Interactions and Model:*  {cloak:id=sysb}[Point As in Part A.Particle Dynamics]{cloak}

{toggle-cloak:id=appb} *Approach:*  

{cloak:id=appb}

Just as above, [Newton's 2nd Law|Newton's Second Law] for the _x_ direction can be written:

{latex}\begin{large}\[ \sum F_{x} = F = ma_{x}\] \end{large}{latex}

This time, however, the acceleration requires some thought.  The speed of the box and its direction of motion are constant.  Thus, by definition, the acceleration is zero.  This impliesFor simplicity, we will refer to the front box as box 2 and the rear box as box 1.  We will call the force applied by box 1 to box 2 _F_~21~.  The only horizontal force experienced by the front box is the force _F_~21~ provided by the rear box.  Thus, [Newton's 2nd Law|Newton's Second Law] for the _x_ direction can be written:

{latex}\begin{large}\[ \sum F_{x} = F_{21} = mam_{2}a_{x} = (\mbox{(1510 kg})(0\mbox{2 m/s}^{2}) = \mbox{020 N} \] \end{large}{latex}

{info}This result is probably not consistent with your everyday experience.  The reason for this is that it is very difficult to find a box and floor combination with zero friction.  Instead, consider the effort that would be required to keep an air-hockey puck moving at constant speed on the air-table (friction is very small) or to keep a soccer ball rolling at constant speed on a smooth, level floor (friction is unimportant since the ball is rolling).{info}

{cloak}cloak}

{card}
{card:label=Part C}

h3. Part C

Suppose we now consider the rear (15 kg) box as a 

{card}
{deck}

{td}
{tr}
{table}
{live-template:RELATE license}