4. Solved Examples
1. A bee strikes a windshield of a car on the freeway and gets crushed. What can you conclude about the force on the bee versus the force on the windshield, and on what principle is this based?
Solution:
The force on the bee is equal in magnitude to the force on the windshield. This is based on Newton's Third Law of Motion: For every action, there is an equal and opposite reaction. While the forces are equal, the effects are very different due to the large difference in masses. The bee experiences a very large acceleration (and is crushed), while the windshield experiences a negligible acceleration.
2. A car is traveling at top speed on the Bonneville salt flats while attempting a land speed record. The tires exert 25 kN of force in the backward direction on the ground. Why backwards? How large are the forces resisting the forward motion of the car, and why?
Solution:
The tires exert a force backward on the ground because, by Newton's Third Law, the ground exerts a force forward on the tires, propelling the car forward. This is the "reaction" force to the tires' "action."
The forces resisting the forward motion of the car are also 25 kN. Since the car is traveling at a constant top speed, its acceleration is zero. By Newton's Second Law (), the net force on the car must be zero. Therefore, the forces resisting motion (air drag, rolling resistance, etc.) must be equal in magnitude to the forward force provided by the tires.
3. A sky diver of mass 90 kg (with suit and gear) is falling at terminal speed. What is the upward force of air drag, and how do you know?
Solution:
The upward force of air drag is equal to the sky diver's weight. At terminal speed, the sky diver's acceleration is zero, so the net force on the sky diver is zero. Therefore, the upward force of air drag must equal the downward force of gravity (weight).
Weight = mg = (90 kg)(9.8 m/s2) = 882 N. So, the upward force of air drag is 882 N. Using g=10, we get 900N.
4. A cart on wheels (assume frictionless) with a mass of 20 kg is pulled rightward with a 50N force. What is its acceleration?
Solution:
Using Newton's Second Law, . Since the motion is only in one dimension, we can write , where Fx = 50 N and m = 20 kg:
5. What will the acceleration be in the previous problem if the wheels effectively give the cart a coefficient of friction of
Solution:
The force of friction is given by , where N is the normal force. In this case, the normal force is equal to the weight of the cart: N = mg = (20 kg)(9.8 m/s2) = 196 N (or 200N using g=10).
So, the frictional force is (or 20N using g=10).
The net force on the cart is now (or 30N using g=10).
The acceleration is (or 1.5m/s/s using g=10).
6. In the previous problem, the rightward pull is now directed at an angle of 30 degrees upward from horizontal while the cart still rolls horizontally. What is the cart's acceleration?
Solution:
The horizontal component of the applied force is .
The upward component of the force reduces the normal force. The net vertical force must be zero, so , which gives (or 200N - 25N = 175N).
The frictional force is (or 17.5N).
The net horizontal force is (or 43.3N - 17.5N = 25.8N)
The acceleration is (or 25.8N/20kg = 1.29m/s/s).
7. Using Newton's 2nd law, show that all objects subject to the pull of gravity alone should fall at the same rate. What is that rate?
Solution:
Newton's Second Law:
The force due to gravity is , where is the acceleration due to gravity.
Setting these equal, . The mass m cancels out, leaving .
Since the acceleration is equal to and is a constant (approximately 9.8 m/s2 or 10m/s/s downward), all objects subject to gravity alone fall at the same rate, which is .
8. What is the third law pair to the normal force as you sit in a chair? What effect does the sun's pull on earth have in terms of third law pairs?
Solution:
The third law pair to the normal force as you sit in a chair is the force you exert on the chair due to your weight. You are pushing down on the chair, and the chair is pushing up on you with an equal and opposite force (the normal force).
The sun's pull on Earth (gravitational force) has a third law pair: the Earth pulling on the Sun with an equal and opposite gravitational force.
9. Two planets, each of mass 1020kg are at coordinates and Find the force on each planet by the other.
Solution:
The gravitational force is given by Newton's Law of Universal Gravitation: , where G = 6.674 × 10-11 N(m/kg)2.
First, find the distance between the planets:
The magnitude of the force is:
Now, find the direction vector. The vector pointing from planet 1 to planet 2 is
To find the unit vector, we divide by the magnitude (which we already have as r):
The force on planet 1 is in the opposite direction to (since the force is toward planet 2). So
The force on planet 1 is
The force on planet 2 is the negative of this.
10. A hydrogen atom has just a single electron orbiting the nucleus, which happens to be a single proton without any neutrons. The proton is positively charged, the electron negatively, but both with the same magnitude of charge given by e=1.602x10-19C. The mass of an electron is 9.11x10-31kg, and the proton is 1.67x10-27kg. Find the ratio of the electrostatic to the gravitational force of attraction between the electron and the proton in hydrogen.
Solution:
The electrostatic force is given by Coulomb's Law: , where k = 8.9875 × 109 N(m/C)2
The gravitational force is given by Newton's Law of Universal Gravitation: , where G = 6.674 × 10-11 N(m/kg)2.
We want the ratio :
. Notice that the distance r between them cancels.
(dimensionless)
11. What is the meaning of a first order approximation?
Solution:
A first-order approximation means approximating a function using its tangent line at a specific point. It's a linear approximation that is accurate near the point of tangency, and it involves ignoring higher-order terms (quadratic, cubic, etc.) in a Taylor series expansion. It is useful for small deviations from the point of approximation.
12. What is a good general rule to follow in order to find the best choice of coordinate system to solve a dynamics problem?
Solution:
A good general rule is to choose a coordinate system that aligns with the direction of the acceleration vector or the direction of the net force. This simplifies the problem by reducing the number of components you need to consider. If acceleration or force is along a curved path, consider using a coordinate system aligned with the tangent and normal directions to the path.
13. A box with friction coefficient of 0.2 rests on a 12 foot long plank of wood. How high (in feet) must one side of the plank be lifted in order for the box to begin to slide?
Solution:
The box will start to slide when the component of gravity parallel to the plank equals the maximum static friction force. Let θ be the angle of the plank. Then:
Now, we can find the height. sin(θ) = height / length, so
height = length * sin(θ) = 12 ft * sin(11.31°) ≈ 2.35 ft.
14. A skateboarder starts from rest and rolls down a 3.0 m long 5% ramp, and then rolls up another 5% ramp. How far will they make it up the second ramp if
a) there is zero friction?
b) there is a friction coefficient of 0.01?
Please do this problem using kinematics and not energy (which you should only know from a previous course, since we have not yet arrived at that topic).
Solution:
First, note that a 5% ramp means that for every 100 meters of horizontal distance, the ramp rises 5 meters vertically. This means so that
a) Zero friction:
The acceleration down the ramp is . The final velocity at the bottom of the first ramp is the final velocity after accelerating at that rate over the 3.0 m.
Using kinematics, so that (the velocity at the bottom).
Coming back up the second ramp the acceleration is the same but in the opposite direction. Thus we want to know how far the skateboarder travels with that initial velocity. So again except now our final velocity is 0 and our initial velocity is 1.71m/s. So so and . The answer is 3.0 meters.
b) Friction coefficient of 0.01:
Now,
Thus, the velocity at the bottom of the first ramp is
Now, on the way up the ramp, the friction will act to slow the skateboard, meaning the new acceleration is
Now so that
15. A toy car speeds up at 1.0 m/s2 while rolling down a ramp, and slows down at a rate of 2.0 m/s2 while rolling up the same ramp. What is the slope of the ramp in degrees? Grade in %? The friction coefficient?
Solution:
Going down,
Going up,
(Note the minus sign, since we are slowing).
Adding the two equations:
Subtracting the two equations:
Grade = tan(θ) * 100% = tan(8.80°) * 100% ≈ 15.4%
16. A crate rests on the back of a flat-bed semi truck that is driving at 20 m/s. The crate is dangerously not strapped down. What is the shortest stopping distance possible for the truck such that the crate does not slide on the truck bed if the friction coefficient between the crate and the bed of the truck is 0.40?
Solution:
The maximum deceleration of the truck without the crate sliding is determined by the maximum static friction force that can act on the crate.
Maximum friction force:
The maximum deceleration of the truck is
Using kinematics to find the stopping distance:
,
where vf = 0 m/s, vi = 20 m/s, and a = -3.92 m/s2
If we use g=10 to make the numbers easier, and the maximum deceleration becomes 4m/s/s. Then from we get .
17. A box is dropped on a level conveyor belt that is moving at 4.5 m/s in the +x direction in a shipping facility. The box/belt friction coefficient is 0.15. For what duration will the box slide on the belt? In which direction does the friction force act on the box? How far will the box have moved horizontally by the time it stops sliding along the belt?
Solution:
The friction force acts on the box in the +x direction, accelerating it until it reaches the belt's speed. The friction force acts on the belt in the -x direction, retarding it very slightly.
The maximum acceleration of the box is determined by the friction force:
(or 1.5 if using g=10).
The time it takes for the box to reach the belt's speed is given by:
vf = vi + at
4. 5 m/s = 0 + (1.47 m/s2) * t
The distance the box travels during this time is:
If we use g=10 to make the numbers easier, the acceleration is now 1.5m/s/s.
The time becomes .
Finally .
18. A little toy car is powered by a CO2 cartridge that produces a thrust force according to the following function: Fthrust/sub = 6.0N e-0.2t. The car also has a friction coefficient of 0.05 due to the wheels and bearings. If the car's mass is 1.2kg, and the car travels in the +x direction, what will it's acceleration function look like as a function of time? If it starts from rest, what will vx/sub be? How fast will the car be going after 3.0s? What maximum speed will be reached by the car?
Solution:
The friction force is . (If g=10, 0.6)
The net force on the car is
The acceleration function is .
To find the velocity function, integrate the acceleration function:
Since the car starts from rest, vx/sub = 0:
0 = -25e0 - 0.49(0) + C
C = 25
So,
The velocity at t = 3.0s is:
To find the maximum speed, we need to find when the acceleration is zero:
The maximum speed is reached at t=12.1s:
If we use g=10 then 0.6 = ma which means there is a constant force of friction of 0.6N. This makes the numbers a bit easier. Now the force due to thrust is 6e^(-0.2t) minus the force due to friction 0.6N, or, 6e^(-0.2t) - 0.6. The object's acceleration will thus be that total force divided by mass which is or . Integrating this function gives velocity where the constants of integration can be simply evaluated to force the initial velocity to 0. Thus, so that at t=3s, . The function for acceleration will be 0 when or or or . The maximum velocity will be
19. A rocket's velocity (measured with respect to a nearby planet) changes from to
upon being struck by an asteroid. The asteroid's mass is 1/10th that of the rocket and upon impact got embedded into the sheet metal of the rocket rather than flying away. What was the asteroid's incoming velocity as measured with respect to the same nearby planet?
Solution:
This problem involves conservation of momentum.
Let mr be the rocket's mass and m[sub]a = 0.1m[sub]r be the asteroid's mass. Let be the initial velocity of the rocket and be the initial velocity of the asteroid, and be the final velocity of the combined rocket and asteroid.
Conservation of momentum:
Divide through by m[sub]r:
20. Two objects get pushed by the same magnitude of force. One object is 10x more massive. How does the rate of change of momentum for the more massive object compare with the less massive one? Please be able to explain why in terms of a quantitative statement found in the chapter.
Solution:
The rate of change of momentum is the same for both objects. Newton's Second Law can be written as , where p is momentum. Since the force is the same for both objects, is the same for both. The momentum will be changing at the same rate.