Chapter 5 Forces in Two Dimensions. Targets I can represent vector quantities both graphically and algebraically. I can use Newton’s laws to analyze motion.

Chapter 5

Forces in Two Dimensions

Targets

• I can represent vector quantities both graphically and algebraically.

• I can use Newton’s laws to analyze motion when friction is involved.

• I can use Newton’s laws with the knowledge of vectors to analyze motion in two dimensions

5.1- Vectors

• Vector Addition 5 N

7 N

What is the total magnitude and direction?

20 N

16 NWhat is the total magnitude and direction?

Vectors in Multiple Dimensions

• Hint: When drawing two-dimensional problems graphically, you need to use a protractor (Please have one by Monday)– Correct angles– Measure the direction and magnitude of the

resultant vector

• Draw vectors tip to tail• Connect the tail of the first vector to the tip of

the last vector– Example #1-

• Use Trigonometry to find the unknown vector’s direction and length– Pythagorean Theorem– Law of Cosines– Law of Sines

Pythagorean Theorem

• Only used when there is a RIGHT ANGLE• A2 + B2= R2 R= Resultant

– Example #2- Find the magnitude of the sum of Chris’s path if he goes south 5 km and then due east 4 km.

Law of Cosines

• R2=A2 + B2 – 2ABcosӨ– Which angles are which

– Example #3- If Connor travels a path that is 15 km and then turns and walks 25 km find the magnitude of the sum when the angle between them is 135o.

• Example #4- Matthew decides to go for a hike. He walks 4.5 km in one direction, then makes a 45o turn to the right and walks another 6.4 km. What is the magnitude of his displacement?

Law of Sines

1

Example 5Law of Sines: a/sin A = b/sin B = c/sin

C

Put in the values we know: a/sin A = 7.0/sin(35°) = c/sin(105°)

Ignore a/sin A (not useful to us): 7.0/sin(35°) = c/sin(105°)

Now we use our algebra skills to rearrange and solve:

Swap sides: c/sin(105°) = 7.0/sin(35°)

Multiply both sides by sin(105°): c = ( 7.0 / sin(35°) ) ×

sin(105°)

Calculate: c = ( 7.0 / 0.574... ) × 0.966...

Calculate: c = 12

Example: Calculate side "c"

 .0

SOH CAH TOA

Component Vectors

• Ax=5• Ay=4

• Components- a vector parallel to the x-axis and another parallel to the y-axis

• Vector resolution- process of breaking a vector into its components

• The direction of a vector is defined as the angle that the vector makes with the x-axis measured counter clockwise

Algebraic Addition of Vectors

• Rx=Ax+Bx+Cx

• Ry=Ay+By+Cy

• Rx and Ry are at 90o angles so the magnitude of the resultant vector can be found by using the Pythagorean Theorem.

• R2=Rx2+ Ry

2

• Angle of the resultant vector Θ=tan-1(Ry/Rx)– The angle of the resultant vector is equal to the

inverse tangent of the quotient of the y-component divided by the x-component of the resultant vector

• Example 6– Add the vectors via the component method. A is

4.0 m south and B is 7.3 m northwest. What is the magnitude of the resultant vector? What is the magnitude of the resultant vector Θ?

5.2- Friction

• Static and Kinetic Friction– Kinetic Friction: exerted on one surface by

another when 2 surfaces rub against each other because one or both of them are moving.

– Static Friction: exerted on one surface by another when there is no motion between the 2 surfaces

• Friction Depend On:– Normal Force applied– Type of Material

– What is the equation for Force of Kinetic Friction?

0 0.5 1 1.5 2 2.5 3 3.50

0.51

1.52

2.53

3.5

Friction

Normal Force

Kine

tic F

ricti

onal

For

ce

Equations

• Kinetic Friction Force– Ffkinetic=µkFN

– The kinetic friction force is equal to the product of the coefficient of the kinetic friction and normal force

• Static Friction Force– Ffstatic≤µsFN

– The static friction force is less than or equal to the product of the coefficient of the static friction and the normal force

• Table 5-1 page 129

• Example 7: Suppose that a particular machine in a factory has two steel pieces that must rub against each other at a constant speed. Before either piece of steel has been treated to reduce friction, the force necessary to get them to perform properly is 5.8 N. After the pieces have been treated with oil, what will be the required force?

• Example 8: A girl exerts a 36 N horizontal force a she pulls a 52 N sled across a cement sidewalk at constant speed. What is the coefficient of kinetic friction between the sidewalk and the metal sled runners? Ignore air resistance

• What is the unit?

• If the object is accelerating, use the equation– Fnet=F-µkFN

• Example 9: You help your mom move a 41 kg bookcase to a different place in the living room. If you push with a force of 65 N and the bookcase accelerates at 0.12 m/s2, what is the coefficient of kinetic friction between the bookcase and the carpet?

5.3- Force and Motion in Two Dimensions

• Equilibrium- net force = zero

• If I have two vectors, how would I find a third vector that brings it to equilibrium?– 1st- Find the sum of the two vectors– 2nd- The equilibriant will be equal in magnitude

and opposite in direction! (Newton’s Third Law)• Equilibriant- a force that puts an object in

equilibrium

Extra Credit

If you do the challenge problem on page 132, show all work, and get the correct answer (showing magnitude and direction), you will receive +2 extra credit points toward your test.

Motion Along an Inclined Plane

• 1st- Identify the forces acting on the object (free body diagram)

• 2nd- Set up your coordinate plane– X-axis= Ff

– Y-axis= FN

• 3rd- Weight will not be equal to the FN so you must break it down into its components.

Example 10

• A crate weighing 562 N is resting on a plane inclined 30.0o above the horizontal. Find the components of the weight forces that are parallel and perpendicular to the plane.