I made a force table that I can use to analyze the component of each force and the net force that results.
Force - y-component
Fg:____-mg
Fa:____+P
Fn:____mg-P?
__________________
ma:____0
Adding the entries in this table, it seems obvious that the normal force must equal mg-P. Otherwise, the...
Why is it I only apply this negative to the normal force? If a downward force is negative, then Fg should be negative too, and
-Fg-Fn+Fa=0
Fn=Fa-Fg=P+mg
Which still isn't the right answer :/
Homework Statement
Let's say I'm holding a crate up against the ceiling with a force of 'P', and the crate is stationary.
What is the normal force?
Homework Equations
Newton's 1st law: ƩF=0
The Attempt at a Solution
I have three forces acting on the crate - The normal force...
My physics instructor and I were arguing about this today. It was a problem involving ranges, but the fundamental issue is this:
If you aim a projectile at a target, and the projectile passes 10 meters below the target, do you aim 10 meters above the target to compensate?
The Problem: An empty capacitor is connected to a 12.0 V battery and charged up. The capacitor is then disconnected from the battery, and a slab of dielectric material (k=2.8) is inserted between the plates. Find the amount by which the potential difference changes, and state whether this change...
I came across this when doing another problem:
Suppose we have 2 numbers, (a+b) and (c+d), which both equal 0.
a+b=0
c+d=0
Then a+b=0=c+d,
Thus, a+b=c+d
However, a+b+c+d=0
Thus, a+b=-c-d
Therefore, a+b=c+d AND a+b=-(c+d)
How is this possible?
It's a rational equation that gives me the velocity of the student.. If I set that equal to the velocity function of the bus, I can find the time where they intersect!
0.17t=(0.085t2+40)/t
t=21.7 s.. Thank you so much!
You mean like so?:
sp(t)=sb(t)
-> v(t)*t=0.085t2+40,
-> v(t) =(0.085t2+40)/t
This tells me what the velocity would be at my intersection point, but I still don't have the time, t.
Homework Statement
This problem is 2.95 of University Physics, 11th edition.
Catching the Bus: A student is running at her top speed of 5.0 m/s to catch a bus, which is stopped at the bus stop. When the student is still 40.0 m from the bus, it starts to pull away, moving with a constant...
Oh wait, the solution just hit me!
For "if ab>0, then either a>0 and b>0, or a<0 and b<0" to be true,
a and b have to ALWAYS be either greater or less than 0. In other words, a and b can never change signs.
Obviously, (x-1) and (x-3) DO change signs at x=1 and x=3 respectively, so the...
If I'm correct, you're saying "if ab>0, then either a>0 and b>0, or a<0 and b<0"
is only true if a and b are functions.
But I considered "a=(x-1) and b=(x-3)". Aren't x-1 and x-3 functions of x?
But I don't understand how the definition, "if ab>0, then either a>0 and b>0, or a<0 and b<0" can remain true in this instance. In other words, I know what to do, but I don't know why it works.
I just got Spivak's calculus today, and I'm already stuck on the prologue problems:
1. The problem
Find all x for which (x-1)(x-3)>0
2. The attempt at a solution
We know that if ab>0, then either a>0 and b>0, or a<0 and b<0.
Thus, if a=(x-1) and b=(x-3), then either (x-1)>0 and...