You are viewing an old version of this page. View the current version.
Compare with Current
View Page History
« Previous
Version 10
Next »
Unable to render embedded object: File (Boxcar and Cannonballs 01.xls) not found.
Imagine that you have an indestructible boxcar sitting on frictionless railroad track. The boxcar has length L, height H, and width W. It has N cannonballs of radius R and mass M stacked up against one end. If I move the cannonballs in any fashion – slowly carrying them, rolling them, firing them out of a cannon – what is the furthest I can move the boxcar along the rails? Which method should I use to move the boxcar the furthest? Assume that the inside walls are perfectly absorbing, so that collisions are perfectly inelastic.
Part A
Solution One
System: Boxcar and cannonballs as point particles.
Interactions: Not Important in this part.
Model: Point Particle Dynamics.
Approach:
Diagrammatic Representation
Unable to render embedded object: File (thatfbd1.jpg) not found.
the system consists of the Boxcar on rails and the Cannonballs, plus whatever devices we use for propulsion inside.There are thus no external influences
Mathematical Representation
Since there are no external influences, which includes forces, the center of mass of the system is not affected, and by the Law of Conservation of momentum must remain fixed. .
Unknown macro: {latex}
\begin
Unknown macro: {large}
[\ M_
Unknown macro: {Boxcar}
x_
Unknown macro: {Boxcar, initial}
+ \sum M_
Unknown macro: {i}
x_
Unknown macro: {i, initial}
= M_
x_
Unknown macro: {Boxcar, final}
+ \sum M_
Unknown macro: {i}
x_
Unknown macro: {i, final}
]\end
Here xi is the position of the center of the *i*th cannonball and xBoxcar is the position of the center of the boxcar. The subscripts initial and final indicate the positions at the start and the end of our operation.
Unknown macro: {latex}
\begin
Unknown macro: {large}
[ N = F_
Unknown macro: {A}
= \mbox
Unknown macro: {300 N}
]\end
Part B
Unable to render embedded object: File (thatnormal2.jpg) not found.
A person moves a 10 kg box up a smooth wall by applying a force of 300 N. The force is applied at an angle of 60° above the horizontal. What is the magnitude of the normal force exerted on the box by the wall?
Solution
System: Box as point particle.
Interactions: External influences from the earth (gravity), the wall (normal force) and the person (applied force).
Model: Point Particle Dynamics.
Approach:
Diagrammatic Representation
We begin with a free body diagram for the box:
Unable to render embedded object: File (thatfbd2.jpg) not found.
Mathematical Representation
From the free body diagram, we can write the equations of Newton's 2nd Law.
Unknown macro: {latex}
\begin
Unknown macro: {large}
[\sum F_
Unknown macro: {x}
= F_
Unknown macro: {A}
\cos\theta - N = ma_
]
[ \sum F_
Unknown macro: {y}
= F_
Unknown macro: {A}
\sin\theta - mg = ma_
]\end
Because Because the box is held against the wall, it has no movement (and no acceleration) in the x direction (ax = 0). Setting ax = 0 in the x direction equation gives:
Unknown macro: {latex}
\begin
Unknown macro: {large}
[ N = F_
Unknown macro: {A}
\cos\theta = \mbox
Unknown macro: {150 N}
]\end
Part C
Unable to render embedded object: File (thatnormal3.jpg) not found.
A person scrapes a 10 kg box along a low, smooth ceiling by applying a force of 300 N at an angle of 30° above the horizontal. What is the magnitude of the normal force exerted on the box by the ceiling?
Solution
System: Box as point particle.
Interactions: External influences from the earth (gravity), the ceiling (normal force) and the person (applied force).
Model: Point Particle Dynamics.
Approach:
Diagrammatic Representation
We begin with a free body diagram for the box:
Unable to render embedded object: File (thatfbd3.jpg) not found.
Mathematical Representation
From the free body diagram, we can write the equations of Newton's 2nd Law.
Unknown macro: {latex}
\begin
Unknown macro: {large}
[\sum F_
Unknown macro: {x}
= F_
Unknown macro: {A}
\cos\theta = ma_
]
[ \sum F_
Unknown macro: {y}
= F_
Unknown macro: {A}
\sin\theta - mg - N = ma_
]\end
Because Because the box is held against the ceiling, it has no movement (and no acceleration) in the y direction (ay = 0). Setting ay = 0 in the y direction equation gives:
Unknown macro: {latex}
\begin
Unknown macro: {large}
[ F_
Unknown macro: {A}
\sin\theta - mg - N = 0 ]\end
which we solve to find:
Unknown macro: {latex}
\begin
Unknown macro: {large}
[ N = F_
Unknown macro: {A}
\sin\theta - mg = \mbox
Unknown macro: {52 N}
]\end