Preparation problems
Handout #3. Wind tunnel measurements show that a
baseball can be suspended nearly motionless in an upward-directed 95-mph
vertical airstream. Thus, the terminal velocity of a baseball is 95 mph
(42.5 m/s). A baseball has a mass of 5 ounces (142 g). (a) If a
baseball experiences a drag force FDrag = Dv2, what is the
value of the drag factor D? (b) What is the magnitude of the
drag force when the baseball is travelling at 80 mph (36 m/s)?
Solution:
(a) We know that terminal velocity is reached when the weight of
the baseball is balanced by the drag force acting on the baseball as it
falls. At this point, when we sum the forces, we find there is no net
acceleration. Thus Newton's Second Law tells us
Further, the net force acting on the baseball is the sum of the drag
force and the weight of the baseball. The drag force is given by
Since we are solving at the condition of terminal velocity, we will
denote this velocity as vt, and look at the net force when that
condition is met
We can solve the above equation to find D, as we know the other values
in the equation. We get
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(0.142 kg)(9.8 m/s2)
(42.5 m/s)2
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(b) We can find the magnitude of the drag force using
Equation (1)
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(7.7 ×10-4 kg/m)(36 m/s)2 |
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Handout #5. A 50-kg parachutist jumps from an airplane
and falls to earth with a drag force proportional to the square of the
velocity. FDrag = Dv2. Take D = 0.2 kg/m (with the
parachute closed) and D = 20 kg/m (with the chute open). (a) Determine
the terminal velocity of the parachutist in both configurations, before
and after the chute is opened.
Solution:
(a) We know that terminal velocity is reached when the
weight of the parachutist is balanced by the drag force acting on the
parachutist as they fall. At this point, when we sum the forces, we find
there is no net acceleration. Thus Newton's Second Law tells us
Further, the net force acting on the parachutist is the sum of the drag
force and the weight of the parachutist. The drag force is given by
Since we are solving at the condition of terminal velocity, we will
denote this velocity as vt, and look at the net force when that
condition is met
We can solve the above equation to find D, as we know the other values
in the equation. We get
Now we can find the terminal velocity in both configurations. First
with the chute closed:
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æ
Ö
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(50 kg)(9.8 m/s2)
0.2 kg/m
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Finally, we can do the same thing for when the chute is open:
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æ
Ö
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(50 kg)(9.8 m/s2)
20 kg/m
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Handout #8. A leading LPGA golfer consistently hits
her 5-iron a distance of 170 yards (155 m). A golf ball, mass 46 g,
experiences a drag force FDrag = Dv2. (a) Using the
observation that a golf ball has a terminal velocity in air of 44.5 m/s,
determine the drag factor D for a golf ball.
Solution:
(a) We know that terminal velocity is reached when the
weight of the parachutist is balanced by the drag force acting on the
parachutist as they fall. At this point, when we sum the forces, we find
there is no net acceleration. Thus Newton's Second Law tells us
Further, the net force acting on the parachutist is the sum of the drag
force and the weight of the parachutist. The drag force is given by
Since we are solving at the condition of terminal velocity, we will
denote this velocity as vt, and look at the net force when that
condition is met
We can solve the above equation to find D, as we know the other values
in the equation. We get
Now we can find the drag factor for the golf ball
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(0.046 kg)(9.8 m/s2)
(44.5 m/s)2
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Follow-up problems
6-15. Your 12-kg baby sister is hanging on the bottom
of the tablecloth with all her weight. In the middle of the table,
60 cm from each edge, is a 6.8-kg roast turkey. (a) What is the
acceleration of the turkey? (b) From the time she starts
pulling, how long do you have to intervene before the turkey goes over
the edge of the table?
Solution:
(a) We begin, as is often the case, by drawing a free-body diagram
for the relavent bodies in the problem.
where W turkey represents the weight of the
turkey, Ntable is the normal force exerted on the
turkey by the table, T is the tension in the tablecloth, and
Wbaby is the weight of the baby. We are trying to
find the acceleration of the turkey. Since the turkey and the baby are
connected by a rope, we see that the turkey will accelerate to the left
(the +x-direction) at exactly the same rate as the baby accelerates
downwards (the -y-direction). So we can write
Now we can write down Newton's Second Law in both the x- and
the y-directions for the baby and the turkey. First we will sum the
forces in the y-direction for the turkey:
This equation does not do us much good, so we move on to sum
the forces in the x-direction for the turkey:
Here we are stuck because we have two unknowns, T and aturkey.
We need another equation to solve the problem, so we look to sum the
forces in the y-direction for the baby:
We can use Equation (2) to get rid of the tension in this
equation. This gives us
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[ mturkey aturkey,x ] - Wbaby |
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[ mturkey aturkey,x ] - (mbabyg) |
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Finally, we can recall from Equation (1) that we can relate the
acceleration of the turkey to the baby. So we substitute in for the
acceleration of the baby to get our answer:
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mturkey aturkey,x - mbabyg |
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mturkey aturkey,x - mbabyg |
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mturkey aturkey,x + mbaby aturkey |
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[mturkey + mbaby] aturkey |
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(12 kg) (9.8 m/s2)
(6.3 kg) + (12 kg)
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(b) Given that we now know the acceleration of the
turkey, it remains a straightforward kinematics problem to determine how
long until the turkey falls off the table. We can use the first
kinematics equation from out Constants and Equation sheet:
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Since we want to know when the turkey reaches the end of the table, I
will say that the turkey is initially at x0 = -0.6 m from the edge of
the table, which I will call x = 0 m. We also know that the turkey
starts at rest, or vx0 = 0 m/s. Solving for t, we get
6-47. During an ice storm, the coefficients of friction
between car tires and road are reduced to mk = 0.088 and
ms = 0.14. (a) What is the maximum slope on which a car can
be parked without sliding? (b) On a slope just steeper than this
maximum, with what acceleration will a car slide down the slope?
Solution:
(a) We begin, as is often the case, by drawing a free-body diagram
for the car on the incline.
where N road is the normal on the car by the
road, W car is the weight of the car, and
Ff is the force of static friction on the car
by the icy road, which lies up the slope, at an angle q to the
horizontal. Since we are trying to determine the maximum angle of the
slope on which the car can remain at rest, we will be in a situation in
which the net force is zero. We will set up our coordinate system as
shown in the figure. This means that we must find the x- and
y-components of the weight. These are:
Now we need to apply Newton's Second Law. Summing
the forces in the y-direction, we see
Now we can write down Newton's Second Law in the
x-direction. This gives us
Now we know from our constants and equation sheet that the maximum force
of static friction is given by Fs = ms N. Substituting
into our above equations, we get
(b) Now we are looking for the acceleration of the car
when the slope is just steeper that that above. If the slope is any
steeper than 7.97°, the car will begin to slip and the frictional
force will no longer be static, but kinetic. Our free body diagram will
look exactly like the one in (a), with the exception that the
frictional force will become Fk.
Despite sliding down the slope, the car will not accelerate
through the road, so the net force in the y-direction will still be
zero. Thus, Equation (1) will still hold. The net force in the
x-direction, however, will result in an acceleration in the
-x-direction, and so Newton's Second Law in the x-direction will
look like:
Since we know the magnitude of Fk = mk N, we can solve
for our only remaining variable, ax once we substitute in for N
using Equation (1):
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(mg) sinq0 - mk (mg) cosq |
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(9.8 m/s2) [ sin(7.97°) - (0.088) cos(7.97°) ] |
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6-59. A police officer investigating an accident
estimates from the damage done that a moving car hit a stationary car at
25 km/h. If the moving car left skid marks 47 m long, and if the
coefficient of kinetic friction is 0.71, what was the initial speed of
the moving car?
Solution:
Since the entire problem takes place on level ground, we can
skip drawing a free-body diagram. We know that the normal force will be
equal to the weight of the car on level ground. We also know that the
frictional force is the only force acting in the horizontal direction,
and that Fk = mk N. Given these facts, we can write down Newton's
Second Law in the horizontal direction:
We can use this acceleration in the third kinematics equation to find
the initial velocity of the moving car. We know the following
information:
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-mk g = -(0.71)(9.8 m/s2) |
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We can put this information into the third kinematics equation to get:
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Ö vx2 - 2 ax (x-x0)
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(6.94 m/s)2 + 2 (0.71)(9.8 m/s2) (47 m)
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Solutions translated from TEX by TTH, version 1.57.
