Good morning. Bobby, please read the problem.
♪ Flipping Physics ♪
An object has the following forces acting on it.
Force 1 equals negative 4.00 i plus 3.00 j newtons. Force 2 equals 1.00 i plus 4.00 j newtons.
Force 3 equals 1.50 i minus 2.50 j newtons. The object experiences an acceleration
of magnitude 2.34 meters per second squared and an initial velocity of 2.50 i meters per
second. Part (a), what is the direction of the acceleration of the object? Part (b),
what is the mass of the object? Part (c), what is the velocity of the object after 15.0 seconds?
Thank you Bobby. Bo, please begin solving the problem.
Sure. … Uh. We can determine the net force acting on the object. That equals … well, plug in all the
force vectors … and combine all the i's and j’s together to get negative 4 plus 1 plus 1.5 all
times i plus 3 plus 4 minus 2.5 all times j. That gives you … negative 1.5 i plus 4.5 j … newtons.
And we could use the Pythagorean Theorem and tangent theta to determine the magnitude
and direction of the net force acting on the object. The Pythagorean Theorem gives us that
the magnitude of the net force equals the square root of the net force in the x-direction squared
plus the net force in the y-direction squared. … Substitute in numbers to get the square root
of negative 1.5 squared plus 4.5 squared or 4.74342 newtons. … And tangent theta equals
opposite over adjacent or the net force in the y direction over the net force in the x direction.
That means theta equals the inverse tangent of the net force in the y-direction divided by the
net force in the x-direction. … Substitute in numbers to get the inverse tangent of 4.5
divided by negative 1.5 or negative 71.5651 or … 71.6 degrees above the negative x-axis.
What does the negative in front of the angle mean? Oh, that just means the angle is clockwise.
When using your calculator and trig functions, positive angles are counterclockwise;
negative angles are clockwise. Right. I forgot about that. Thanks.
You are welcome. But, we have not answered
any of the questions yet. - Yeah. - Oh.
Right. This is all great information, however, how are we going to use it to
answer the questions in the problem? … We could use Newton’s Second Law.
(Sure Bobby. How is Newton’s Second Law going to help us here?)
Well, Newton’s Second Law states that net force equals mass times acceleration.
Force and acceleration are both vectors and mass is a scalar, therefore, the net force
acting on the object and the acceleration of the object are in the same direction. …
Actually, that means we already have the answer for the direction of the acceleration of the
object. 71.6 degrees above the negative x-axis. Very nice Bobby.
Oh, and we can also rearrange Newton’s Second Law to solve for the mass of the object.
It equals the net force divided by acceleration. We already have the magnitude of the net force
and the magnitude of the acceleration was given in the problem. So, 4.74342 divided by 2.34
equals 2.027101 or 2.03 kg with 3 significant digits.
Thank you Billy. … We have solved parts a and b of the problem. The last part is part (c).
Bo, how do you think we can find the velocity of the object at 15.0 seconds?
Let’s see. … We know the acceleration has a magnitude of 2.34 meters per second
squared. That means the acceleration is not changing and we can use the uniformly accelerated
motion equations. … Let’s use final velocity equals initial velocity plus acceleration times
time. We have all those values so … Velocity at 15 seconds equals the initial velocity or v naught
or 2.5 i plus the acceleration or 2.34 times the time which is 15 seconds. … That works out to be …
Wait a second. We can’t do that can we? (Do what Billy?)
… Combine the unit vector form of a vector and the magnitude of a vector in the same equation.
Nice catch Billy. No, we cannot do this. You have to either have everything in terms of unit vectors
or make sure all of your vectors are broken into components in perpendicular
directions and deal with the perpendicular directions separately using magnitudes.
Oh yeah, the acceleration has a magnitude of 2.34 meters per second squared
and is at an angle of 71.6 degrees above the negative x-axis, however, the initial velocity
is only in the positive x-direction. We cannot combine those into one equation.
Unless we use unit vectors. Right, unless we use unit vectors.
So, let’s do that. Billy, how do we determine the acceleration in terms of unit vectors?
Okay. … We go back to using Newton’s Second Law again. Only now we solve for acceleration and get
acceleration equals net force divided by mass. If we plug in the unit vector form of the net force,
we will get the unit vector form of the acceleration. Cool! Okay, … negative 1.5 i plus
4.5 j divided by 2.027101 equals negative 0.739973 i plus 2.21992 j meters per second squared.
Thank you Billy. Bo, please finish determining the velocity of the object at 15 seconds.
Sure. Velocity at 15 seconds equals the initial velocity or 2.5 i plus the acceleration or
negative 0.739973 i plus 2.21992 j times the time which is 15 seconds. … That works out
to be … 2.5 i plus negative 11.0996 i plus 33.2988 j or, after combining the i values
and rounding to 3 significant digits, negative 8.60 i plus 33.3 j meters per second.
Thank you Bo. I want to take the time (Uh, mr.p?)
Yes Billy? Part (c) asks for the
velocity of the object. Don’t we have to determine the magnitude and direction of the velocity
by using the Pythagorean Theorem and SOH CAH TOA? Ah, good question Billy. No, we do not need to.
The velocity we have determined is in its unit vector form which is an
equally valid way to describe the velocity of the object. We certainly could determine the magnitude
and direction of the velocity, however, the question does not ask us to, so we do not need to.
Okay. Thank you mr.p. You are welcome Billy. … Okay, I
want to take the time to discuss the directions of the velocity and the acceleration of the object.
We already know the net force acting on the object and the acceleration of the object
are in the same direction. Are the velocity and acceleration of the object in the same direction?
… No, the initial velocity of the object is in the positive x-direction and the acceleration is
directed at 71.6 degrees above the negative x-axis. Those are not the same direction.
If the initial velocity of the object were zero, then the directions of the acceleration and
velocity of the object would always be the same. Right, but that’s not what is happening here.
Actually, in this case, the net force acting on the object will cause an
acceleration on the object which will push the direction of the velocity of the object
toward the direction of the acceleration, however, they will never quite end up being in the same direction.
Very nice everybody. Thank you very much for learning with me today,
I enjoyed learning with you. … Actually, one more quick thing. … I do want to point out
that the net force we solved for which acts on the object is, in fact, the sum of the 3 forces. We
can show that using tip-to-tail vector addition. … We can take force 1 … and add force 3 to it,
because the order does not matter, … and then add force 2. … And you can see that the net
force acting on the object is 4.74 newtons at 71.6 degrees above the negative x-axis
or, if you prefer the unit vector version, negative 1.5 i plus 4.5 j newtons.
Cool. Right. I’m done … now.