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Physics “Magic Trick” on an Incline

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– Good morning. Today I am going to perform my one and only magic trick. – Really Mr. P.? ♫ Flipping Physics ♫ – Absolutely, but first,
we need to do some physics. – Sure, that makes sense. [Mr. P.] We start by making
sure the table is level, we do this so we can determine
the angle of this incline. The length of the board
is 70.5 centimeters and the top end of the
board is 15.9 centimeters above the table. Bo,
please use this information to determine the incline angle. – Sure, sine of theta equals
opposite over hypotenuse and opposite the incline
angle is 15.9 centimeters and the hypotenuse is 70.5 centimeters taking the inverse sine
of the whole equation gives us the incline angle
equals the inverse sine of the quantity 15.9 divided by 70.5, which is 13.0342 degrees. – Now that we have the incline angle, Bobby could you please
tell me all the forces that are currently acting on this block? – The force normal is up and
perpendicular to the incline, and the force of static
friction is parallel to and up the incline, and the force
of gravity is straight down. – [mr.p] Bobby, thank you for showing
that you should always draw the free body diagram before
breaking forces into components and then redraw the free body diagram
with the components. Now, use the components
of the force of gravity. – Okay, rather than the force of gravity, we have the force of
gravity perpendicular, which is perpendicular
to the incline and down, and the force of gravity parallel, which is parallel to and down the incline. – Let’s now use Newton’s
second law to determine the magnitudes of all these forces. Oh, actually we need the
mass of the block first, which is 121 grams, which we
can multiply by one kilogram over 1000 grams to get 0.121 kilograms, the mass of the block. Bobby, Newton’s second law please. – The net force in the y, no wait, the net force in the
perpendicular direction equals positive force normal minus
force of gravity perpendicular which equals mass times
acceleration in the perpendicualr direction, and the
acceleration of the block is 0 in all directions
because it isn’t moving, so the force normal equals the
force of gravity perpendicular and we have an equation for the force of gravity perpendicular,
it equals mass times the acceleration of
gravity times the cosine of theta which then is 0.121
times 9.81 times the cosine of 13.0342, which is 1.15643, or 1.16 newtons with three sig figs. – Notice how the force normal
does not equal the force of gravity which means
it does not equal mass times the acceleration due to gravity. I know, many times before
the force normal was equal to mass times the
acceleration due to gravity, however that is clearly not always true. In order to find the force normal, you have to draw a free body
diagram and sum the forces. Billy, the parallel direction please. – The net force in the
parallel direction equals the force of static
friction, which is positive because it is up the
incline, and the force of the gravity parallel
which is negative because it is down the incline,
which then equals the mass of the block times its acceleration
in the parallel direction. Bobby already pointed out the acceleration in all directions of the block is zero, so the force of static
friction equals the force of gravity parallel, which is
equal to the mass of the object times the acceleration due to gravity, times the sine of the incline angle, which is 0.121 times 9.81
times the sine of 13.0342, which is 0.267709 or 0.268 newtons with three significant figures. – At this point, we have
determined the magnitudes for all the forces which
are currently acting on th e block, and everything we’ve done so far is typical of the way we
solve physics problems. What I am about to do,
is to solve for numbers for the magic trick. Again, this portion of what
we’re doing is for the magic trick only, and you would only do this
part if you are going to do this magic trick for your
friends and family at home, not when solving a
physics problem, got it? – Sure. – Yes. – Okay, I’ll take notes then. – Very nice, Bo. The mass in the parallel direction equals mass times the sine of the incline angle or 121 grams times the sine of 13.0342, which is 27.2894 or with
three sig figs 27.3 grams. The mass in the perpendicular direction equals mass times the
cosine of the incline angle or 121 grams times the cosine of 13.0342, which is 117.883, with
three sig figs, 118 grams. Now, I need a few more
things for the magic trick. You can see I have added a
string in the perpendicular direction and a string
in the parallel direction and each string goes over a pulley and is attached to a 5 gram mass holder. Now the reason I figured
out our mass’ parallel and perpendicular in grams is because these masses, which I
can add to the 5 gram mass holders come in gram increments. Okay, let’s start with
the parallel direction, we have 5 grams right there, I need to get to 27.3 grams or 27 because I only have gram increments, so 27 grams is what we’re going for in the parallel direction. 5 grams, if we add 20 to that, we now have 25 grams and if we add 2 more
grams, we now have 27 grams in the parallel direction.
Now let’s do the perpendicular direction, you can see we need 118 grams in the perpendicular direction, we already have 5 grams
in the 5 gram mass holder adding 100 grams gets us to 105 grams in the perpendicular direction, adding 10 more gets us to 115 grams in the perpendicular
direction, 2 more is 117 grams, and 1 more gram gets us to 118 grams in the perpendicular direction. Now, it might not seem
like I have done much until I do this. (applause and chatter) – Ladies and gentle
people, I present to you a floating block. Okay,
to be completely honest with you it’s not magic, it’s physics, so who can explain to me what I did? I took the board away
and yet the block remains exactly where it was before
I took the board away, so why is that? Yes, Bobby? – You replaced the forces
the board was applying on the block, with tension
forces in the string. – Yeah, if you multiply the mass parallel and perpendicular in
kilograms by the acceleration due to gravity, you get the
magnitudes of the forces we solved for before in the
parallel and perpendicular directions. – Right, so the force normal and the force of static friction
are each replaced by tension forces, and voilà, the
board wasn’t applying any forces on the block anymore, so you could remove it. [Mr. P.] Exactly, the floating
block ends up being a great visual for the directions of
the forces acting on the block, which is in equilibrium. The
force normal from the board which is now a force of
tension, was perpendicular to the incline and up, the
force of static friction, which is now a second force of tension, was parallel to and up the incline, and the force of gravity is straight down. Thank you very much for
learning with me today, I enjoyed learning with you.

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