Unit 2, Part 1: Forces in One Dimension
Velocity and Acceleration fall under a category of physics called “Kinematics”. Kinematics explains “how” things move. For example, “That car is moving 5m/s to the right” or, “That car’s velocity is changing from 10 m/s to 5 m/s because of the negative acceleration” are both statements that would describe how something is moving, but not why.
We’re now about to study a category of physics called “Dynamics”. Dynamics explains “why” objects move. For example, dynamics will explain why objects in the air fall down and not up. Or, dynamics will explain why a car speeds up or slows down or why you can push a box more easily on ice than you can on cement.
This unit is all about forces. A force is simply a push or a pull. Forces CAUSE objects to accelerate (or objects to stretch, bend, squeeze, etc). So, a force will cause an object’s velocity (it’s speed and/or its direction) to change. Forces are vector quantities, so “force” has BOTH size and direction (we’ll use positive or negative for direction). The direction of an object’s acceleration (change in velocity) by a force is always in same direction as the force.
Forces CAUSE masses to accelerate. So, an overall net force will cause an object’s velocity to change from some initial velocity to some final velocity. The equation that will show just what size of force will produce what acceleration is
Fnet = ma Note: also written as SF = ma, read as “the sum of the forces acting on an object is equal to the object’s mass multiplied by its acceleration”
This equation says that a net overall Force of size “F” will cause a mass of size “m” to accelerate with an acceleration of “a”. The net Force that causes the acceleration is the mathematical sum of all the forces acting on an object. So, if you take ALL the forces acting on an object and add them together, THAT is the force that will cause the object to accelerate. If the NET overall force is zero, there will be NO acceleration. The acceleration is always in the same direction as the NET force acting on the object.
The unit of Force is called a “Newton” and is abbreviated with the symbol “N”.
Forces ALWAYS occur in pairs. If I “push” you on the arm, your arm “pushes” back on my hand with a force that is equal in size but oppositely directed. So, if I push on your arm with a force of 10N to the right, your arm will push back on my hand with a force of 10N to the left. Notice that the size of the force (the “10N”) is equal for both, but the direction is opposite (left vs. right).
Do the following example problems. Make sure you know that the “Force” these problems refer to is the overall “Net” force acting on the objects in the problems. These four problems will teach you how to do basic problems with “Fnet = ma”. Notice in the examples that the unit of force (Newton) results from the mathematical product of a “Kg” and “m/s2 ”. So, 1N = (1Kg)(1m/s2). This says that “one Newton of force” will cause a mass of “one Kg” to be accelerated at a rate of “one m/s2”. In order to make sure that your math comes out correct in “Force” problems, all your units should be in meters, seconds, and kilograms.
Problems for you to practice with.
1. A 1500 Kg car is moving with an initial velocity of 20 m/s and comes to a stop in 5 seconds. What is the size of the net force acting on the car to get it stop in this amount of time?
m = 1500 Kg Fnet = ma need “a” first a = ?
Vi = 20m/s
Vf = 0m/s Fnet = (1500 Kg) x (-4 /s2) a = -4 m/s2
t = 5 s
Fnet= ? Fnet = -6000 Kg m/s2 -6.00 x 103 N
The negative sign of your answer says the force is directed in the opposite direction of the forward motion of the car. If the force was positive, the acceleration would have been positive and the car would have sped up. Since the car was clearly losing velocity (slowing down), the acceleration had to be negative and the force must have been negative, too.
2. When a shot-putter exerts a net force of 140N on a shot, the shot has an acceleration of 19 m/s/s. What is the mass of the shot?
Fnet = 140N Fnet = ma
a = 19m/s2 m =
m= ? m = 140 (kg)(m/s2)
19m/s2
m = 7.37 kg
*Notice in the previous problem that I broke the unit Newton (N) down into the units that make up a Newton in order to make sure everything cancels.
3. Together a motorcycle and rider have a mass of 275 Kg. The cycle is slowed with an acceleration of –4.5 m/s/s. What is the net force on the bike? Describe the direction of the force and the meaning of the negative sign.
m= 275kg Fnet = ma
a = -4.5 m/s2
Fnet = ?
4. A 1225 Kg car traveling at 105 Km/h slows to a stop in 53m. What is the size and direction of the net force that acted on the car?
m = 1225kg Fnet = ma -----à you need “a” first
d = 53m
vf = 0m/s Vf2 = Vi2 + 2ad --à solve for “a”
vi = 105 km/h ?m/s
Fnet =?
The six main forces that we’ll be speaking about in here (these make up the NET force) are: KNOW THESE!!!!!
FA = The “Applied Force”. This is the force that some person is applying to an object by pushing or pulling it.
Ff = The “Friction Force”. The friction force always opposes the motion of two objects sliding against each other (always acts opposite the direction the object is moving). Sometimes the FA might equal the Ff , but sometimes they won’t. That will be part of the “problem” for you to figure out.
Weight (mg) =This is the “Force of Weight”. This is the mathematical product of your mass (in Kg) and gravitational acceleration (9.8 m/s2). It is ALWAYS directed down, straight toward the center of the earth. (Fw = mg ; the symbol we will use for weight in this class is “mg”…more on that later)
N = The “Normal Force”. This is the force that any surface pushes back on an object that is pressed against it (such as the ground pushing back up on an object that is on the ground). This force only acts on objects that are pressed against something (like the ground, or a table).
FL = The “Lift Force”. This is some upward force that causes an object to move “up” in the air (like the engines on the space shuttle).
T = The “Tension Force”. This is the force in a rope that pulls an object up through the air (like a spider or a string) or along a surface (like a rope pulling a sled along the ground).
The following diagram represents all the forces acting on an object that is on the ground or on some other horizontal surface. The object can be a box, a car, a wheelbarrow, a person, etc. If it’s on a surface, these are the ONLY FOUR forces we’ll be concerned with. The directions in which the forces act are given by the direction of the arrows.
· Things sliding along a horizontal surface (such as a box sliding along the floor) with no additional upward or downward force acting on them will always have the size of the weight (mg) equal the size of the N so they will always cancel each other out. If there is an additional force pressing down along with the weight of the box itself (like if someone came and sat on the box), the normal force will increase due to the normal force always equaling the total force that is pressing the object into the surface. If someone is pulling up on the object (with a rope, etc) the normal force will become less than the actual weight of the box itself. You can’t control an object’s weight---it is only determined by an object’s mass and gravitational acceleration. However, you can control the normal force by pulling up on or pushing down on an object that is in contact with the floor.
· If the FA is greater than Ff, then the overall “net” force along the surface the object is sliding across is in the direction of the FA and the object will accelerate in that direction. If Ff is greater than FA, then the Fnet will be in the direction of the Ff and the object will accelerate in that direction.
N
Ff FA
mg
The following diagram represents the only two forces (there are only two) that act on an object that is in the air (like a balloon or a helicopter) but not in free-fall. You can tell it isn’t in free-fall by the upward force acting on the object in addition to the downward force of its weight. The only force acting on objects in freefall is the object’s weight (mg). If FL is greater than mg then the object will accelerate upward, and vice-versa. Notice there isn’t an applied or friction force on this object. Objects in the air aren’t in contact with a surface, so there won’t be any friction (or applied force).
FL
mg
Or, if an object is hanging by a line, the only two forces acting on it will look like the following diagram. If the T > mg, the object will accelerate upward and vice-versa.
T
mg
WEIGHT AS A FORCE
These next problems have to do with the force your body exerts on the earth. This force is called “weight”. Your weight is a force that always is directed down, straight down toward the center of the earth. The “force” of weight is the product of an object’s mass and the “acceleration due to gravity” (mg). Gravitational acceleration varies on different places on Earth (it decreases as you get further away from the center of Earth) and varies on other moons/planets. The moon is much smaller than Earth, so its gravitational acceleration (g) is much smaller than Earth (more about this in a future chapter). Therefore, your weight (mg) will be much smaller on the moon than Earth.
Fweight = (mass) x (gravitational acceleration --- 9.8 m/s2) Fw = mg
5. What is the weight of each of the following objects?
a. 0.113 hockey puck b. 108 Kg person c. 870 Kg car
Fw = ?
m = 0.113 kg m = 108kg m = 870 kg
a = 9.8 m/s2
force of weight = ma = mg
6. Find the mass of each of the following objects:
a. 98N b. 80N c. 0.98N
m = ?
g = 9.8 m/s2
weight = mg= 98N mg = 80N mg =0.98N
*Remember: The units that make up a Newton are “kg m/s2”. Use these units in place of the “N” to do your unit cancellation (and check your answers!).
7. A 20N stone rests on a table. What is the force the table exerts on the stone? In what direction?
Answer: The force the stone exerts on the table is 20N down (mg = 20N). Since the stone is NOT moving, the acceleration MUST be zero (things that aren’t moving don’t have “changing velocity”, so the “a” is always “zero”). If the acceleration is zero, the Fnet must also be zero (since Fnet = ma). Therefore, the 20N down must be balanced by 20N up. The upward force comes from the force the table exerts on the stone (the normal force). This is Newton’s 3rd Law. Those three laws are coming up shortly in the notes…..
8. Suppose Joe, who weighs 650 N, stands on a bathroom scale calibrated in Newtons.
a. What force would the scale exert on Joe? In what direction?
b. If Joe now holds a 50N cat in his arms, what force would the scale exert on him now?
c. After Joe puts the cat down, his dad comes up behind him and lifts up on his elbows with a 100N force. What force does the scale exert now?
Fw = 650N
a. Since he isn’t accelerating up or down, his Fnet in the “up and down” direction must be zero. The only forces on him in the up and down direction are his mg (pushing down on the scale) and the N (the force the scale pushes up on him). Therefore, the upward force of the scale pushing UP on him (N) is equal in size but opposite in direction to the downward force he exerts on the scale.
So, FN = 650N
b. Fw = (650N) + (50N) = 700N, down
FN = 700N, up
c. Whatever force is pressing DOWN on the scale (total downward force), the scale will push back UP with the same value (N). Joe is pressing down on the scale with mg = 650N. But his father comes up and pulls UP on him with an upward force of 100N. Therefore, the TOTAL downward force is actually the difference between the two forces[650N, down –100N, up = 550N, down). Since this force is directed DOWN, and the scale pushes back UP with 550N of force.
9. An astronaut with mass 75 Kg travels to Mars. What is his weight on
a. Earth?
b. Mars (where g = 3.8m/s/s)?
c. What is the value of g on top of a mountain if the astronaut weighs 683N there?
a. m = 75kg Fw =ma
aearth = 9.8 m/s2
mg(earth) = ?
b. mg(mars) = ?
a = 3.8 m/s2
m = 75 kg
c. g = ?
mg= 683N
m = 75 kg
To understand pretty much EVERYTHING in this class the rest of the year, you need to KNOW and UNDERSTAND Newton’s Three Laws of Motion. These laws explain the relationship between “acceleration” and its cause… “Force”. Remember that if an object is accelerating its velocity is changing. So, things that aren’t moving OR are moving with some constant velocity (10 m/s, for example) aren’t accelerating., so their Fnet must equal zero.