I have tried to google this but don't know what to google. I know that normal force is the force exerted back on an object that opposes gravity, perpendicular to the surface. My question is, what are the physics of an object that isn't just flat on a surface, but only a fraction of it is. In other words, an object sitting partially off of a cliff. At what point does the force of gravity take over and make it fall as you slowly slide it towards the edge.
Newtonian Mechanics – Object on a Ledge
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Related Solutions
In the example of the incline plane that you have provided, the mass does not affect the speed, because the only friction force present is proportional to the object's weight. However, oftentimes significant dissipative forces are proportional to the velocity of the object --- for example, if an object if freely falling through a viscous fluid. You could model such situation by the equation below:
$$ F = ma = mg - kv $$
Here, the coefficient $k$ of the drag force is some parameter that could depend, for example, on the object's shape. We cannot simply divide through by $m$ to solve for $a$; moreover, we have a differential equation, since $a$ is the derivative of $v$. If you solve it, you would indeed get that $v$ is a function that depends on mass. We first notice that when the object reaches the velocity large enough for the drag force to cancel $mg$ completely, the object will maintain that velocity, since there will be no acceleration. If we substitute $a = 0$ in the equation, we get that this terminal velocity should equal to $v_{terminal} = \frac{mg}{k}$. We could actually solve this differential equation by using a substitution function $u = g - \frac{k}{m}v$ and the fact that $a = v'$. We get then:
$$ u' = -\frac{k}{m}u $$
Guessing a solution $u = C_{1}e^{-\frac{k}{m}t} + C_{2}$, where $C_1$ and $C_2$ are some constants, we get that
$$ v(t) = \frac{mg}{k} - C'e^{-\frac{k}{m}t} - C'' $$
Where some constants $C'$ and $C''$ depend on our initial conditions; if the initial velocity of the object is zero, the function that would work is:
$$ v(t) = \frac{mg}{k}(1 - e^{-\frac{k}{m}t}) $$
Where we set $C' = \frac{mg}{k}$ and $C''= 0$, so that the boundary conditions $v_{initial} = 0$ and $v_{terminal} = \frac{mg}{k}$ are met. We ended up with a velocity function that depends on time and the mass of the object.
- As for the second question: let's track back to when we set the object in motion. We know that a coefficient for static friction is greater that the coefficient of kinetic friction: $\mu_{static} > \mu_{kinetic}$. This ensures that objects resting on rough services don't start moving from a slightest touch, but rather require an initial "jerk". Hence the initial force with which we will be pushing an object to start it moving, $F_{initial}$, will be greater than the eventual force $F_{final}$, to provide that "jerk": $F_{initial}$ is slightly greater than $\mu_{static}mg$, and $F_{final}$ is just equal to $\mu_{kinetic}mg$ to ensure constant velocity. We only need to exert $F_{initial}$ for a short amount of time to get the object going; but in that short time, it accelerates, and the magnitude of the acceleration depends on the force: the larger the mass of the object is, the larger the static friction is, and the larger $F_{initial}$ has to be. That implies a greater acceleration in the same amount of time --- and hence a greater constant velocity.
Whenever you're confused about forces and work, you can bring it back to energy. The real definition of work is the transfer of energy. If no energy is transferred, no work is done. Ever. Force times distance doesn't define work, it quantfies it. The only mechanism for the transfer of energy is force exerted through a distance.
But there doesn't have to be a nonzero net force. If you push a box across a floor against a friction force at constant velocity, both you and the friction force are doing work. Some people would say that equal amounts of positive and negative work were being done on the box. I don't like that. I prefer to say that chemical energy from you is being transferred to thermal energy in the box and the floor. That keeps it real.
In the case of an object or system traveling toward the center of the Earth at constant velocity, energy is being transferred by the force of gravity from the gravitational field to the person or thing applying the equal and opposite force.
Best Answer
If you have learned about free body diagrams, you need to create one, but instead of adding one normal force, add two forces, spanning the base of the object
Now you need to balance the weight $W$ with the two normal forces $N_1$, $N_2$ in the vertical direction, as well as balance the moments about the center of mass (the summation point is arbitrary here).
This leads to two equations, with two unknowns, the two normal forces
$$\begin{aligned} N_1 + N_2 & = W \\ b\,N_1 -a\,N_2 & = 0 \end{aligned}$$
The condition for stability is that $\boxed{N_2 \ge 0}$. Otherwise, you need to push down on the back of the object to keep it from going over.