Jhaj, David asks:

CATAGORY: Mechanics QUESTION: An airplane travels 80m/s as it makes a horizontal circular >turn which has a 0.80-km radius. What is the magnitude of the resultant >force on the 75- kg pilot of this airplane (in kN)?

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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


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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