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Study Of Motion in Two Dimensions

Movement in Two Dimensions: This page focuses on several different scenarios involving motion in two dimensions.

Equilibrium and the Equilibrant:

As was said in the study of forces, an object is in a sate of equilibrium with respect to forces when the net force acting on it equals zero. [F(net) = 0 Newtons, in the metric system, or 0 lbs, in the English system.) When an object is in this state of equilibrium, it is either standing still with respect to a reference point or it moving with constant velocity with respect to a reference point. As we have already seen, an object that is in a state of equilibrium will NOT experience an acceleration.

Consider the diagram below. If Vector A and Vector B are added the sum is the resultant. The Equilibrant is the vector that is equal in magnitude to the Resultant, but opposite in direction. When the three vectors, Vector A, B and C are added the sum is 0 Newtons and the object is in a state of equilibrium with respect to the forces acting on it. The object is either at rest or moving with constant velocity. According to Newton's laws of motion and object in a state of equilibrium would not experience any acceleration since the F(net) would be zero newtons. To Review the information about vectors and adding vectors go back to the student information page and click on the link to the "Power Point Notes and Slides". This will provide you access to the power point presentations about vectors and adding vectors.

Now consider a situation which focuses on creating an equilibrium situation. Study the diagram first and then move on to the example problem which follows. Forces F(A) and F(B) are supporting the sign, while the Force of Gravity is pulling down on the sign.

[Note: Forces acting at right angles to one another do not affect one another.] A 200 N sign is hanging motionless from two cables that are both 35 degrees from the horizontal. What is the Tension in the cables? G: F(g) = W = 200 N; Both angles are 35 degrees. And because of the congruency of the two triangles F(A) and F(B) are equal. F: Tension in the two cables, when they are equal angles from the horizontal. Remember vector addition in two dimensions is not the same as simple arithmetic addition such as is used for vector addition in one dimension. In this problem the following statements apply. F(A) = F(Ay) + F(Ax) F(B) = F(By) + F(Bx) R = F(Ay) + F(By) | R | = | F(g) | = 200 N F(Ay) = F(By) = 100 N F(net,x) = 0 Newtons; F(net,y) = 0 Newtons The horizonatal forces; E: F(net,x) = F(Bx) - F(Ax);      F(net,x) = F(B) * cos angle - F(A) * cos angle S: And Since 0 Newtons = F(B) * cos angle - F(A) * cos angle A: It follows that F(B) * cos angle = F(A) * cos angle While the horizontal forces do cancel each other out, they have no effect in the vertical direction and contribute nothing to holding up the sign. They do,however, keep the sign in the center position. The vertical forces; E: F(net,y) = F(Ay) + F(By) - F(g)     F(net,y) = F(A) * sin angle + F(B) * sin angle - F(g)     and since F(Ay) = F(By)     then 2 F(A) * sin angle = F(net,y) + F(g)     and F(A) =  ( F(net,y) + F(g) ) / (2 * sin angle) S: F(A) =  ( 0 N + 200 N) / ( 2 * sin 35 ) A: F(A) = 174 N; and because the tension is the same in both cables     F(B) = 174 N

Inclined Plane: An inclined plane also known as a ramp is a commonly employed structure which allows movement in both the horizontal direction and the vertical direction at the same time.  We often see them used on trucks to help load and unload objects.  They are very useful for getting wheeled vehicles up and down from one level of elevation to another.  Two common applications of the use of the inclined plane are wheel chair access ramps and the ramps used in multilevel parking garages.

The study of inclined planes requires that we focus on two important forces. One force is the normal force and the other is the parallel force.  The normal force, which is always perpendicular to the surface upon which the object is found, can be described as the force that acts against or pushes against the object.  This force could be considered to be the convenient force, because it results from the surface's ability to support the weight of the object either resting or moving across the surface.  The ability of the surface to support the object without collapsing and allowing the object to fall is dependent upon both the materials and the design used in constructing the surface.

On horizontal surfaces the magnitude of the normal force is equal to the magnitude of the force of gravity acting on the object.  The force of gravity acting on an object is measured as the weight of the object.  In the British (English) system of weights and measures the commonly used unit of weight is the pound (lb.).  In the metric system it is the Newton (N).  On an inclined plane the magnitude of the normal force is less than the magnitude of the force of gravity acting on the object.  As the slope of the inclined plane becomes steeper the value of the normal force decreases.  As A, the angle of inclination, increases from 0 to 90 degrees, the normal force decreases in value.  The equation for finding the normal force is:  F(normal) = F(N) = Fg cos A .

On inclined planes which are tilted at some angle measured from the horizontal there is a force called a parallel force.  In a situation where there is no friction this parallel force would definitely cause the object to accelerate down the inclined plane.  The steeper the inclined plane the faster the object would accelerate.  When the surface is horizontal meaning the angle of inclination is 0 degrees, the parallel force is 0 N.  When the surface is vertical meaning the angle of inclination is 90 degrees, the parallel force will equal the force of gravity (Fg, the weight of the object.).  The equation for finding the parallel force is: F(parallel) = F(p) = Fg sin A .

Now when there is friction the parallel force would be acting downwards along the ramp while the force of friction would be opposing that motion.  In this situation the relationship among the forces can be expressed by the following equation F(net) = F(p) + F(f) .  In many instances an additional force, an applied force, is present and that would increase the complexity of the equation.  One form of the equation could be written as F(net) = F(p) + F(f) + F(A) . Below, we have an example problem involving an inclined plane in a frictionless environment.

 Example:  In a frictionless environment a 10 kg mass is placed on an inclined plane that is tilted 30 degrees from the horizontal. Calculate: a. The force of gravity acting on the object. b. The normal force acting on the object. c. The parallel force acting on the object. d. The acceleration moving the object down the incline. Part a solution: G: m = 10 kg, angle of inclination = 30 degrees (from the horizontal) F: The force of gravity acting on the object (the weight of the object) E: Fg = W = m g S: Fg = 10 kg * 9.8 m/s^2 A: Fg = 98.0 N Part b solution: G: The above information F: The normal force E: F(N) = Fg cos A . S: F(N) = 98.0 N * cos 30 A: F(N) = 84.87 N Part c solution: G: The above information F: The parallel force E: Fp = Fg * sin A . S: Fp = 98.0 N * sin 30 A: Fp = 49.0 N Part d solution: G: The above information F: The acceleration of the object E: a = Fp / m S: a = 49.0 N / 10 kg A: a = 4.9 m/s^2

Projectile Motion: This is the curved motion of an object that is projected into the air.  The trajectory of a projectile is the name we give to the path followed by a projectile.  A projectile can be thrown sideways (horizontally) or thrown up at some angle measured from the horizontal.  The study of projectile motion describes an object's motion along its trajectory. The shape of a trajectory depends on the observer.

Imagine a passenger, riding on a bus, moving with a constant velocity, tossing a ball up and down.  From the perspective of the passenger the ball is simply going up and down.  We can refer to this observation as that of the passenger.  The passenger would be using the bus, or the floor of the bus as his or her reference point (also called a frame of reference).  The ball rises and falls back to its starting point.  Now ask yourself how would the motion of the ball appear to someone standing by the side of the road as the bus passes by.  The ball would now appear to follow a parabolic curve rising and falling while moving sideways much like a ball appears to do when tossed from one person to another.  This view sees the person tossing the ball and the ball both moving sideways, but the ball is also rising and then falling while moving sideways.  The person standing by the side of the road would be using the ground as their frame of reference.  Events often look different when perceived from different points of view (in different frames of reference).

Horizontally launched projectiles: These are often called cliff problems because throwing an object sideways off of a cliff is a classical example of this kind of problem.  The time in the air is affected by the height at which the object begins its flight and the acceleration caused by gravity (g).  To solve this kind of problem you need to know how to use several common equations used in the general study of motion. The equations usually associated with this sort of problem are:

To find the time to fall to earth, d(vertical) = d(y) = v(y) t + 0.5 g t^2, and t = [2d(y) / g]^0.5

To find the horizontal distance or range, d(horizontal) = d(x) = v(x) t

To find the final vertical velocity just before impact, v(y,final) = v(y,initial) + g t

Also remember that horizontal vector quantities are independent of vertical vector quantities.  The horizontal velocity in this sort of situation is independent of the vertical velocity except for the fact that they are linked by the time interval controlling the flight of the object.

In a frictionless environment the horizontal velocity is constant while the vertical velocity is constantly changing because of gravity.  What this means is that when a projectile such as a 0.22 caliber bullet is fired horizontally (sideways), it will fall towards the ground at the same rate as a 0.22 caliber bullet simply dropped simultaneously from the same height. Example: A stone is thrown horizontally at 15 m/s from the top of a cliff that is 44 m high.  Answer the following questions assuming no friction. a. How long does it take the stone to reach the ground? b. How far from the base of the cliff does the stone strike the ground? Part a solution: G: The initial vertical velocity = 0 m/s, The vertical distance = 44 m, and the initial horizontal velocity is 15 m/s F: The time to fall to the ground E: d(y) = v(y)*t + 0.5 g t2 , becomes t = [ 2 d(y)  / g ]^0.5 S: t = ( 2 * -44 m / -9.8 m/s^2 )^0.5 A: t = 3.0 s Part b solution: G: info in part a F: The horizontal distance traveled by the stone as it falls E: dx = vx t S: dx = 15 m/s * 3 s A: dx = 45 m

Consider this hypothetical question.  Suppose a hunter looks across a valley and sees an animal a considerable distance across the valley from him at the same height as he is as measured from the valley floor.  He sights his rifle to point directly at the animal and shoots.  If the animal could see and think about what is happening should it jump off the side of the valley and fall downward when the rifle is fired or should it remain where it is in order to have the hunter miss striking him?  If the hunter is concerned with hitting the game animal, should he do just as has been described or should he aim differently?  Explain.

Also imagine a ball thrown sideways.  As it moves forward it also falls downwards.  If you kept up with it moving horizontally yourself, the ball would simply appear to fall downwards.

Vertically launched projectiles: This type of motion could be illustrated by throwing a ball from one person to another.  The velocity of a projectile is resolved into horizontal and vertical components.  When solving problems, the horizontal and vertical velocities are treated separately.  The time in the air is affected by the vertical component of the initial velocity with which the object is launched, the angle of inclination measured from the horizontal, and the value of g.  The equations usually associated with this sort of problem are:

To find the vertical component velocity, v(y) = v sin A, where v is the initial or launch velocity.

To find the horizontal component velocity, v(x) = v cos A, where v is the initial or launch velocity.

To find the total time the projectile is in the air, t = [v(y,f) - v(y,i)] / g

To find the horizontal distance or the range, d(x) = v(x) t

To find the maximum height, d(y) = v(y) t + 0.5 g t^2, where the t is 1/2 of the total time Example: A golf ball is struck and leaves the tee with a velocity of 25 m/s at a 35 degree angle measured from the horizontal.  Answer the following. a. What is the vertical velocity of the ball? b. What is the horizontal velocity of the ball? c. How long will the ball remain in the air? d. What is the horizontal distance the ball will travel while in the air? e. What is the maximum height reached by the golf ball? Part a solution: G: initial velocity = 25 m/s, angle of elevation is 35 degrees F: The vertical velocity of the ball (how fast it rises upwards) E: v(y) = v sin A . S: v(y) = 25 m/s * sin 35 A: v(y) = 14.34 m/s Part b solution: G: above information F: The horizontal velocity of the ball (how fast it moves sideways) E: v(x) = vi cos A . S: v(x) = 25 m/s * cos 35 A: v(x) = 20.48 m/s Part c solution: G: above information F: The total time the ball will remain in the air E: t = [v(y,f) - v(y,i)] / g S: t = (-14.34 m/s  14.34 m/s) / (-9.8 m/s^2), REM: v(y,i) = + 14.34 m/s, v(y,f) = -14.34 m/s A: t = 2.927 s Part d solution: G: above information F: Horizontal distance traveled by the ball while flying through the air E: d(x) = v(x) t S: d = 20.48 m/s * 2.927 s, REM: the total time is used because it affects the total horizontal distance traveled. A: d = 59.94 m Solution part e: G: above information F: Maximum height E: d(y) = v(y) t + 0.5 g t^2 S: d(y) = 14.34 m/s (2.927 s / 2) + 0.5 (-9.80 m/s^2) (2.927 s / 2)^2 A: d(y) = 10.49 m

Uniform Circular Motion: This type of motion occurs when a net force acts at right angles to an object moving at a constant speed.  The net force causes the object to experience an acceleration at 90 degrees to its forward motion resulting in the object following a circular path if the net force remains constant.  The use of the word uniform implies that the acceleration is constant and it therefore follows that the net force producing the uniform acceleration is constant (uniform).

The force that pulls the object towards the center of the circular path is called the centripetal force.  The acceleration produced by this force is called the centripetal acceleration.  The length of the path is the Circumference, and the time to travel once around the circular path is called the period.  The word period is used for a time measurement whenever a situation involves a repetitive (cyclical) action.  Each time an object completes one trip around the circular path it has completed one cycle.

Another measurement called frequency is also associated with circular motion.  The frequency is a measure of how many times a cycle is repeated per second (cycles per second).  The unit for frequency is the Hertz abbreviated Hz.  If an object were to complete 5 complete trips around a circular path in one second, it would have a frequency of 5 Hz.  Frequency is related to the period of an object experiencing cyclical motion.  The period is equal to the reciprocal of the frequency.  So the object just described has a period equal to one over 5 Hz.  This would mean that the period is 0.2 or 1/5 seconds per cycle.

Imagine that you are on an amusement park ride that is designed as a circular room or platform.  Each person must stand such that they have their back against the wall. As the room or platform spins the rider feels like they are being pressed against the wall as the floor drops from beneath them.  The sensation while the ride spins is experienced as a force pressing them against the wall preventing them from sliding down the wall when the floor is lowered. This perceived force is called the centrifugal force.  It is not a real force at all.  It is the result of your moving body trying to move forward along a straight line tangent to the curve of the wall while at the same time the wall is pushing you towards the center of the ride as the room or platform turns.  There is no actual force pushing you outwards. The only force actually acting on you is the centripetal force.

The equations associated with solving these type of problems are:

To calculate the period of the moving object, T = 1 / f, where frequency is expressed in Hz

To calculate the circumference of the path, d = C = 2 (pi) (radius)

To calculate the velocity of the object, v = C / T

To calculate the centripetal acceleration experienced by the object, a(c) = v^2 / r

To calculate the centripetal force acting on the object, F(c) = m a(c) Example: A 12 kg object is moving around a horizontal circular path with a frequency of 0.5 Hz.  The path has a diameter of 3.0 m.  Do the following tasks. a. Calculate the period of the moving object. b. Determine the circumference of the path. c. Find the velocity of the object. d. Compute the centripetal acceleration experienced by the object. e. Derive the centripetal force acting on the object. Part a solution: G: A 12 kg object moving in a circular path, diameter = 3.0 m, frequency = 0.5 Hz F: Period of the moving object E: T = 1 / f S: T = 1 / 0.5 Hz   [REM: 1 Hz = 1 cycle / second] A: T = 2 seconds / cycle  OR simply 2 seconds Part b solution: G: Information above F: circumference of the circular path E: C = 2 (pi) r S: C = 2 * 3.14 * 1.5 m A: C = 9.42 m Part c solution: G: Information above F: The velocity of the object E: V = C / T S: V = 9.42 m / 2 sec A: V = 4.71 m/s Part d solution: G: Information above F: The centripetal acceleration E: a(c) = v^2 / r S: a(c) = (4.71 m/s)^2 / 1.5 m A: a(c) = 14.79 m/s^2 Part e solution: G: Information above F: The centripetal force E: F(c) = m a(c) . S: F(c) = 12 kg * 14.79 m/s^2 A: F(c) = 177.5 N

Torque: In studying circular motion one looks at examples of objects circling a central point as they follow a curved path.  When the centripetal force is constant, the shape of the path will be circular.  When studying such motion the objects mass is treated as a point mass.  That is, the mass of the object is considered to be present at a specific point in space rather than spread out occupying a defined volume in space.

To explain forces applied to rigid objects such as wrenches that are pivoted about a central point rather than revolving around some centrally located object or point in space, the subject of torque is introduced into the study of force and motion.  A device that lends itself to the study of torque is the simple seesaw as seen on some playgrounds.  A child sits on one end while a second child sits on the other end.  One child rises into the air while the other descends to the ground.  There is a pivot point between them allowing one end of the bar or plank they are sitting on to rise while the other descends.  If they are of similar mass, they will be able to make each other go up and down by pushing up with their legs when they touch the ground on their end.  In simple terms a seesaw is a balance beam where masses of objects are compared with one another from one side of the balance point to the other.  If either side has more mass than the other, the inequality of the weight (a measure of force) from one side to the other will result in a rotation involving the heavier object on the one side moving downwards while the lighter object on the other side moves upwards.  The addition of an applied force on the end with more weight can make the heavier side rise while the end with less weight descends.  This is what often happens when there is some small difference in weight between two children riding on the seesaw.  The heavier child can push harder with their legs than the lighter child does in order to get themselves into the up position and the lighter child into the down position.  The lighter child only has to take advantage of the heavier childs weight to get to the up position.

In the example above, we know that a balance between the heavier child and the lighter child can be achieved if the distance between the heavier child and the pivot point is made to be less than the distance between the lighter child and the pivot point.  That is true, because what must be equal on each side is a measurement called the torque.  By definition torque is the math product of the force applied to a lever arm and the length of the lever arm.  The force can be something as simple as the weight of a child.  The lever arms length is the distance between the pivot point and the point at which the force is being applied to the arm.  From a mathematics point of view, if the sum of the two torque values is equal to zero, the children on the seesaw will be balanced.  Since torque is the product of force and the length of the torque arm, the unit for torque is the Nm. Example:  A wrench is being used to tighten a bolt.  Suppose a 10 N force was applied to the wrench handle 25 cm from the pivot point.  Calculate the torque being applied by the wrench to the bolt. G:  F = 10 N,  torque arm length (or L) = 25 cm = 0.25 m F:  The torque being applied by the wrench to the bolt E:  torque = F * L S:  torque = 10 N * 0.25 m A: torque = 2.5 Nm