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Trigonometric Functions

Deals with the relationship among the angles and sides of a right triangle. SOH CAH TOA

sin Ө=

cos Ө=

Tan Ө=

hypotenuse

sideopposite

hypotenusesideadjacent

sideadjacentsideopposite

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Trig Functions (cont.)When talking about opposite and adjacent sides, refer back to the angle we are working with.

ΘAdjacent Side

Opposite

Side

Hypotenuse

Θ

Opposite Side

Adjacent

Side

Hypotenuse

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Adding Perpendicular Vectors

Perpendicular vectors can be added using the Pythagorean Theorem (a2 + b2 = c2)

Example: An airplane is flying east at 90.0 km/h. The wind is blowing north at 50.0 km/h. What is the resultant velocity of the plane?

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Vector ComponentsWhen a vector is not on one of our axes, it is helpful to think of it as a resultant of two perpendicular vectors. These perpendicular vectors are called its components. The components are usually set up as an x-component and a y-component.

To determine the components of a vector, make a right triangle out of it and use SOH CAH TOA. Example: Find the x- and y-components of a 15 m/s velocity @ 30°.

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Analytical Method of Vector Addition

Refers to an algebraic/trigonometric way to add vectors together.

Remember the graphical way that we added vectors was the head-to-tail method.

The idea behind the method is to add x-components to x-components and y-components to y-components.

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Analytical Method of Vector Addition (Process)• Identify each vector to be added.• Break down each vector into an x- and

a y-component.• Make 2 lists: one of all x-components

and one of all y-components.• Add all of your x-components together.• Add all of your y-components together.• Combine the x-component sum and the y-

component sum using the Pythagorean theorem.

• Use SOH CAH TOA to find the direction of the resultant.

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In-Class Problems (II)

1. Add the following vectors together: A. 50 m/s at 25 degrees southeast B. 24 m/s at 40 degress southwest C. 70 m/s at 32 degress northwest

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In-Class Problems (II)

2. A sign is held up by two ropes as shown in the picture below. What net force are the ropes applying on the sign?

21 N 16 N

80° 60°

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In Class Problem (II)

3. Winston, Deuce and Dexter pull on a 0.5kg chew toy as shown in the diagram below. What is the net force and acceleration on the chew toy?

7 N @ 90°32 N @ 50° 15 N @ 0°

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EquilibriumEquilibrium occurs when the net force acting on an object is zero. When in equilibrium, the object has no acceleration.

Example: A tug-of-war match is in equilibrium when both sides pull with the same force.

You may be asked to find the force required to put an object into equilibrium. This force would be equal to the resultant, but in the opposite direction.

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In class problem (II)

4. A 33 N force to the north and a 44 N force at 60 ｰ northeast act on a boat. What is the magnitude and direction of a third force that would keep the boat from moving (equilibrium)?

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Equilibrium Pre-Lab

Purpose Practice adding vectors using the analytical method of vector addition.

Experimentally find the equilibrant required to put the system into equilibrium.

Materials: Force Table 4 sets of slotted weights

Each set includes - 20g hanger, 4-20g, 4-10g, 2-5 g = 150 g.

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Equilibrium Pre-Lab

Pre-Lab Calculations 1. Find the force required to put the following system into equilibrium:

0.98 N at 60° NW 1.47 N at 70° SW

2. Find the force required to put the following system in equilibrium:

0.98 N due East 0.98 N due North 0.98 N at 20° NW

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Equilibrium Pre-Lab

Procedures: For trial 1, set up the given force vectors on the force table.

Place the desired force on the third hanger and move it around the force table until the ring is centered over the post.

Repeat for the other trials.

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Equilibrium Pre-Lab

Data: Forces Equilibrant

Trial 1: 0.98 N(100g) North 1.37N (140 g) @

0.98 N (100g) East ______________

Trial 2: 1.47 N (150 g) W 0.98 N (100g) @ 0.196 N (20g) @ 45° NE ______________ 0.588 N (60g) @ 45° SE

Trail 3: 0.735 N (75 g) @ 22° SW ___________ 0.98 N (100g) @ 70° SE @ 73.8°NE

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Equilibrium Pre-Lab

Data: Forces Equilibrant

Trial 4: 0.49 N (50g) @ 25° SE________________

0.2445 N (25g) @40° SW @ 88 °NE 0.686 N (70 g) @ 32° NW

Trail 5: Prelab Calculation 0.98 N due East Use answer from Prelab 0.98 N due North 0.98 N at 20° NW

Trial 6: 0.98 N (100g) East Use answer from Prelab 0.98 N (100g) North 0.98 N (100g) @ 20° NW

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Components on an Inclined Plane

When finding vector components on an inclined plane, it is often helpful to adjust our axes. Because all motion will be up or down the incline we will use this as our x-axis. We call this our parallel axis.

Because our normal force is perpendicular to our inclined surface, we will use this as our y-axis.We call this our perpendicular axis.

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Components on an Inclined Plane (II)

Example: Find the components of the weight of a 100 N box that is sitting on a 20° inclined plane.

20°

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In-Class Problems (II)

5. What are the parallel and perpendicular components of a 100kg trunk on an inclined plane that is 21 degrees from horizontal? What is the acceleration of the trunk if the incline plane is frictionless? (a = 3.51 m/s2)

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In Class Problem (II)

6. A 562-N trunk is on a frictionless plane inclined at 30.0 from the horizontal.

Find the parallel and perpendicular components of the weight. (F = 486.7 N, F = 281 N)

Find the acceleration of the trunk down the inclined plane. (a = 4.9 m/s2)

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7.

What happens to the parallel and perpendicular forces when the angle of the inclined plane increase? What happens to the acceleration of the object? What component determines the acceleration of an object?

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Independence of Motion in Two Dimensions

The horizontal and vertical velocities of a projectile have no effect on each other. A projectile launched horizontally will hit the ground at the same time as a projectile that was dropped from the same height.

The horizontal velocity of an object will remain constant even though the vertical velocity continues to increase.

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Objects Launched Horizontally

The initial velocity of the projectile is the velocity in the x-direction. This velocity is constant until the projectile hits the ground.

The initial velocity in the y-direction is 0 m/s. This velocity increases just as it does when the projectile is dropped.

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8. Batman is driving the Batmobile with a velocity of 50m/s when he goes over the edge of a cliff. It takes the Batmobile 3 seconds to convert to a jet.

If the cliff is 75 meters high, will the Batmobile turn into a jet before it hits the ground? (yes, t = 3.91 s)

What will Batman’s vertical velocity be as it hits the ground?

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9.

Mr. Abel jumps horizontally off of a high dive platform. The height of the platform is 10 m. If he travels a horizontal distance of 5 meters before hitting the water, what was his initial velocity? (Vyi = 0 m/s, Vxi = 3.5 m/s)

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Objects Launched at an Angle

The first task is to determine the initial x- and y-components of the velocity using SOH CAH TOA.The x-component will stay constant until the projectile hits the ground.The y-component will decrease on the way up, be zero at the highest point, and increase on the way down. This is the same way the velocity behaved for an object that was thrown straight up in the last chapter.

The projectile will reach its highest point halfway through the total time of flight.

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10.

A cannon shoots a cannonball with an initial velocity of 20m/s at an angle 60° above the horizontal.

How long is the cannon in the air? (t = 3.53s)

If the cannon is aimed at a castle 40 meters away, will the cannonball hit the castle? (no, dx = 35.3 m)

What is the maximum height the cannon ball will reach? (dy = 45.93m)

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11.

Justin throws a javelin a distance of 50 meters. If the javelin was in the air for 3 seconds, find the overall initial velocity (magnitude and direction) of the javelin.

(Hint: You know that vy = 0 m/s after 1.5 seconds because this is at the top of the javelin’s trajectory. This should allow you to find viy.) (Vxi = 16.7 m/s, Vyi = 14.7 m/s, Vyf = 22.2 m/s @ 41.4° NE )