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I will focus here on what the pilot wil feel.
If the plane were not traveling in a circle but in a straight line there
would be the force of gravity to contend with. Gravity pulls down on the
pilot and the normal force of his chair pushes up on him. To the pilot it
would feel like he was being pushed up by a force equal to his weight
(mg). Now say the plane stays level but makes a turn. Then the side of
the plane away from the center of the circle pushes against him as the
circle is executed with a force equal to the centrepital force mv^2/r.
Because both gravity and the centrepital force are orthogonal the
magnitude of the force is given by the square root (sqrt) of
the squares (here a^2 means a times a) of each part.
(1) F = sqrt((mg)^2 + (mv^2/r)^2)
now mg = 75kg X 9.8m/s^2 = 735N =.735kN
and
mv^2/r = 75kg X (80m/s)^2 / (.8km X 1000m/km) = 600N =.600kN
Substituting these values into (1) gives us
(2) F = sqrt((.735)^2 + (.600)^2) = .949kN ~.950kN
***********************
* *
* F = .950kN *
* *
***********************
CATAGORY: Mechanics
QUESTION: An 8.0 kg object rests on the floor of an elevator which is
accelerating downward at a rate of 1.3 m/s-squared. What is the
magnitude of the force the object exerts on the floor of the elevator?(in
Newtons)
The object is decending so the acceleration is a = -1.3 m/s^2. There is a
force (ma) on the object that produces the acceleration and is the sum of
all the external forces on the object. In this case the downward force of
gravity (-mg where g=9.8m/s^2) and the upward normal force of the floor on
the object (N) which is the reaction of the force of the object on the
floor. The magnitude of N then is the size of the force the object exerts
on the floor. Thus we have
(1) ma = -mg + N
Solving for N gives
(2) N = ma + mg
Substitution of m=8.0kg, a =-1.3m/s^2, g = 9.8m/s^2 gives
(3) N = 8.0kg(-1.3m/s^2) + 8.0kg(9.8m/s^2) = 68N
************
* *
* N = 68N *
* *
************
CATAGORY: Mechanics
QUESTION: A 30-kg child rides on a circus Ferris wheel that takes her
around a vertical cirrcular path with a radius of 20m every s. What is
the magnitude of the resultant force on the child at the highest point on
this trajectory (in kN)?
This is an example of a BOTCHED question. It happens every now and then
but sometimes we can learn something anyway. The FIRST problem is that
the rotation period is given as 1s(!) Try to imagine someone on a
circular Ferris wheel that has a radius of 20m (~60ft) going around every
second! Not a pretty sight! That is 125.6m/s = 281.1 miles per hour! OK
so that is probably a typo. The next problem is that they ask for the
resultant force. Since the child is traveling on a circular path the
mafnitude of the resultant force on the child is always equal to the
centrepital force (mv^2/r where v is the tangential velocity, r is the
radius and m is the mass of the child) and this will be true regardless of
the childs position on the wheel. What was probably meant was What is the
force of the child against the seat. In other words if a scale is put
under the child how much would the scale register at the top of the
wheel. In that case we have to keep track of all the external forces that
act on the child that literaly force her to move on the circular path.
The two forces will be gravity and the normal force of the seat on the
child. So in words we have
(1) centrepital force = net external forces on the child
or
(2) centrepital force = gravity + normal force of seat on child
In mathematics we have
(3) mv^2/r = -mg + N
Where m = mass of the child, g = 9.8m/s^2, N =normal force of seat on
child.
By action and reaction the magnitude of N is the push of the child against
the seat so we need to solve for N
(4) N = mv^2/r + mg
IF we had the correct period (lets call it T) then v =2 Pi r /T where Pi
=3.14159.... and we could substitutie into (4) to give a formula
for N.
(5) N = m(2 Pi r /T)^2/r + mg
This is as far as we can take it without having the proper numbers.
CATAGORY: Mechanics
QUESTION: Two blocks of same mass (m) hang on one side of a pulley
attached to a ceiling and on the other side there is a block with mass
3m. If m=3.0 kg, what is the tension in the string connecting the two
objects of same mass (in N)? Assume that all surfaces are frictionless.
A classic pulley and tension problem. Here is a diagram of the system.
____________ celing
|
(+) pulley
| |
| | T(1)
3m [_]|
| | ^ ^
[_] 1m | g = - g z
| T(2) V
[_] 1m
We will take the downward direction to be the negative direction. Because
the left side is heavier than the right side the left side goes down and
the right side goes up. To find the tension T(1) and T(2) we need to
consider "free body diagrams" of the 3m mass and the 1m masses.
Free body diagrams of the 3m and 1m weights
T(1) T(1) T(2)
^ ^ ^
| | |
| | |
3m [_] 1m [_] 1m [_]
| || |
| || |
v vv v
3mg 1mg T(2) 1mg
3m upper 1m lower 1m
mass mass mass
The net force on the 3m mass given by mass times acceleration. If we call
the acceleration -a (the minus indicates down) then for the 3m mass
(1) 3m(-a) = T(1) - 3mg or
-3ma = T(1) - 3mg
The net force on the first 1m mass is given by
(2) 1ma = T(1) - 1mg - T(2)
Note here that we have a instead of -a because the 1m mass is going up.
T(2) pulls down on this mass so it is included as a negative T(2).
On the lowest 1m mass we have
(3) 1ma = T(2) - 1mg
Here T(2) is pulling up on the lowest weight so it does not
have a minus sign. We now have three equations in three
unknowns a, T(1) and T(2).
(4) -3ma = T(1) - 3mg
1ma = T(1) - 1mg - T(2)
1ma = T(2) - 1mg
Solve this problem by substitution and elimination. From the last
equation solve for a.
(5) a = T(2)/1m - g
Substitute for a in the first two equations.
(6) -3m(T(2)/1m - g) = T(1) - 3mg
1m(T(2)/1m - g) = T(1) - 1mg - T(2)
These reduce to
(6) -3T(2) + 3mg = T(1) - 3mg
T(2) - 1mg = T(1) - 1mg - T(2)
Which further reduces to
(7) 3T(2) + T(1) = 6mg
2T(2) = T(1)
Substitution of the second equation into the first for
T(1) in (7) gives
(8) 3T(2) + 2T(2) = 6mg or
5T(2) = 6mg
6
T(2) = - mg
5
And since T(1) is equal to 2T(2)
(9) 12
T(1) = -- mg
5
with m = 3.0 kg , g =9.8 m/s^2 we have
******************
* *
* T(1) = 70.56 N *
* T(2) = 35.28 N *<--- this is the tension between the
* * 1m masses.
******************
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