Vectors 2 Applying the Basics

Vectors 2 Applying the Basics

Rope Pulling Problems:
Assume no friction for now 8.0N 4.0 Kg 35.00 What is the acceleration of the box along the floor as it is pulled?

Understanding the problem without math: 8.0N 4.0 Kg 35.00 What two directions is he pulling as he pulls on the rope? RIGHT & UP Which way is the crate actually accelerating? To the Right!

Understanding the problem without math cont.: 8.0N 4.0 Kg 35.00 Only the Right component of this force is moving the crate! (The up component is wasted force) So only part of the 8.0 N on the rope will move the crate- part of it is wasted.

Understanding the problem without math cont.: 8.0N 4.0 Kg 35.00 F=ma So why isn’t the answer 2.0 m/s2? Not all of the 8.0 was to the right! Which component of the force should we look for first? The right component that moved it to the right!

Now the Math… hyp 8.0N 4.0 Kg 35.00 ? (6.6 N) Which Function? Now what?... adj Cos θ = adj/hyp Cos = adj/8.0N Cross-multiply! Cal= R= 6.6 N right (east)

Assume no friction for now 8.0N 4.0 Kg 35.00 (6.6 N) How much acceleration will the box have along the floor as it is pulled? Cal= 1.65 6.6 N F=ma or… a= R= 1.7 m/s2 right a= F/m 4.0 kg

Hanging objects problems:
How much tension would be on the ropes in each situation below? 25.0 ? ? ? ? 88N 111N 111N on each 222N 125N 88N What direction? Why add .0 to 25?... Each part should have the same number of sig. figs. as the whole

Hanging objects problems:
How much tension would be on the ropes in each situation below? 25.0 ? ? ? ? 88N 111N 111N on each 222N 125N 88N The Point: If an object is not moving, the net force must = 0. The total vectors UP must be = to the weight down.

A B C 480 Fw=99 N How much tension is on each metal rod?

Note that A has left and up components.
B C 480 Note that A has left and up components. 33N 33N 33N 33N 33N 33N 99/3= 33N 99N= Fw Fw=99 N A Which rod will not be the same as the others? Rem, The total vectors UP must be = to the weight down.

Careful with that algebra Eugene!
480 What function? 33N 33N hyp 33N Note that the hypotenuse is the actual rod! opp 480 θ? Sin θ = Opp/Hyp Careful with that algebra Eugene! Fw=99 N What is the tension on B and C? 33N What is the tension on A?

A B C 480 The algebra! 33N 33N Sin θ = Opp/Hyp hyp 33N (44N) opp 33N Sin 480 = 480 hyp 1 Cross multiply! Fw=99 N = (33N) (1) (Sin 480)(Hyp) Cal= (Sin 480) (Sin 480) R= 44N

Why is there more force on A than on B and C? 33N 33N 33N (44N) opp
480 Why is there more force on A than on B and C? 33N 33N 33N (44N) opp 480 A had wasted force to the left! Fw=99 N It still had to have exactly 33N up so that total up would = total down. clip..\..\combo.vob

Incline Plane Problems:
Assume frictionless for now What is the acceleration as the crate slides down the incline? 525 N 30.00

Consider the problem first without math… 525 N 30.00 What force is making it slide? Gravity! Will the acceleration be 9.80 m/s2? No! It is not in freefall

Cont… Down! In what direction is weight always? 525 N 30.00 525 N

This component of weight acts perpendicular to the surface.
Force perp. Cont… 30.00 525 N A Portion of this weight force will press the crate against the incline’s surface. This component of weight acts perpendicular to the surface. We will call it the force perpendicular.

This component of weight acts parallel to the surface.
Cont… 30.00 525 N Force parallel A portion of this weight force acts down the incline, which makes it slide in that direction. This component of weight acts parallel to the surface. We will call it the force parallel.

525 N Cont… 30.00 525 N Which force do you think we will solve for to find the acceleration down the incline? force parallel Now for the math…

525 N Cont… 525 N The Ground angle and the angle θ shown
30.00 30.00 525 N The Ground angle and the angle θ shown will always be the same. Want to know why? Yes NO

All three angles in a triangle must add up to_______ 1800
Cont… ?0 60.00 θ 30.00 90.00 30.00 All three angles in a triangle must add up to_______ 1800 Consider this right triangle… This angle is obviously 900 (a right angle) What must the 3rd angle be? 60.00! ( = 120.0… )

Now note the following Right triangle…
Cont… 60.00 θ 30.00 90.00 30.00 Now note the following Right triangle… All Right angle are 900 total… If this much of the right angle is 60.0… What must θ be? Thus the ground angle will = θ 30.00! ( ? = 90.00)

Which side from θ are we looking for? The Opposite side.
Cont… θ 30.00 30.00 525 N hyp opp Which side from θ are we looking for? The Opposite side. Which side is the Force of weight? The hypotenuse Which function? Sin θ = Opp/Hyp

525 N Cont… Mass of crate 525 N hyp Fw=ma Sin θ = Opp/Hyp m= F/a opp
30.00 30.00 Mass of crate 525 N hyp Fw=ma 263 N Sin θ = Opp/Hyp m= F/a opp Opp Sin = m= 525N/9.80 m/s2 525N 1 m= 53.6 kg Cal= 262.5 R= 263 N (II) down the incline

What is the acceleration as the crate slides down the incline? opp
Cont… θ 30.00 30.00 525 N hyp What is the acceleration as the crate slides down the incline? opp F=ma a=F/m F=263 N (II) down the incline a= 263N/53.6 Kg Cal= m= 53.6 kg 4.91 m/s2 down incline clip1..\..\ clip2..\..\

Vector mini II: Hanging objects, incline planes, and rope pulling problems
1) Jane pulls spot in a wagon. The Mass of the wagon and spot together is 14.4 kg. If she pulls the handle with 44N at a ground angle, how much will the dog and wagon accelerate? 2) A 4.40 Kg crate slides down a incline. What will the acceleration be? A B 3) θ1 θ 2 θ1 =45.00 θ2 =55.50 Find the Tension on rod A and B: 5.00 Kg

River Problems:

River Problems: cont. Clip for after..\..\o.vob
Which takes the most time…? A The North-East movement B The Eastern movement C The Northern movement THE TIMES ARE THE SAME!

River Problems: A boat travels due east at 8.88 m/s. There is also a 2.22 m/s current south acting on the boat. The river is m wide. A) How long will it take the boat to cross the river? B) How far downstream will it end up from where it started?

Here’s both vectors: 1 at a time.
Both affect the boats movement like this… motor’s influence: 8.88 m/s Current’s influence: 2.22 m/s- (note it is slower moving)

Viewing #1: motor’s influence Current’s influence
Boat’s actual direction

Boat’s actual direction
#2 Watch again carefully, does the current seem to effect the time it takes to cross? motor’s influence Current’s influence Boat’s actual direction No, all three take the same time.

Part A: A boat travels due east at 8.88 m/s. There is also a 2.22 m/s current acting on the boat. The river is m wide. A) How long will it take the boat to cross the river? Always be sure that the vectors you use in a formula are in the same direction!

The river is 55.5 m wide- what direction is this measurement?
motor’s influence: 8.88 m/s Current’s influence: 2.22 m/s Boat’s actual direction Which velocity represents the boat movement in that direction?

55.5 m wide motor’s influence: 8.88 m/s v= d/t d= 55.5 m t= d/v v= 8.88 m/s t= ? t= 55.5 m Cal= 6.25 8.88 m/s R= 6.25 sec.

Part B Which velocity vector carries the boat downstream?
A boat travels due east at 8.88 m/s. There is also a 2.22 m/s current acting on the boat. The river is m wide. B) How far downstream will it end up from where it started? Which velocity vector carries the boat downstream?

motor’s influence Current’s influence 2.2m/s Boat’s actual direction

Which movement took 6.25 sec?
motor’s influence 8.88 m/s 6.25 sec! Current’s influence 2.22 m/s Boat’s actual direction All three take the same time.

Note all vectors are in the same direction (up/down)
d=? (the distance downstream) t= 6.25 sec (for all vectors) V= 2.22 m/s (the current downstream) v= d/t d= vt d= (2.22 m/s) (6.25 sec) R= 13.9 m downstream Cal=

Non-900 problems (honors)
What if vectors are not 900 apart… which is very often the case in real life. For Example. (DO not write any of this in your notes- just watch and ask questions- the real sample problem will follow.) A 44 N Force west and a 55N force at 350 act on an object. 44N 55N What is the resultant force?

Non-900 problems 55N 44N They must be at right angles to do this!
Note they are not 900 apart… So you cannot add them head to tail as you normally would. 55N 44N They must be at right angles to do this!

Non-900 problems What to do then? First let’s consider it without math… Break each individual vector into its right angle components (NSEW). 55N is both east and north. 55N 44N 44 N west is already due west. You can leave it alone.

Non-900 problems They exactly oppose each other. 55N
East and west now can be subtracted… Why subtract them? They exactly oppose each other. 55N 55N is both east and north. 44N The net force (in this case west) Can now be combined with the northward component vector left.

Non-900 problems Net east/west 55N 55N is both east and north. 44N
so, add the remaining vectors to get the final magnitude and direction. Now it’s a basic 2 vector problem. Find the Magnitude and direction of the resultant!

Try one… As a boat travels 3.33 m/s due south, there is a current pushing 2.22 m/s toward What is the resultant direction and speed of the boat?

Can they be added head to tail?
As a boat travels 3.33 m/s due south, there is a current pushing 2.22 m/s toward What is the resultant direction and speed of the boat? 90O = 50.00 Can they be added head to tail? (from 0) 1800 0O θ? 50.00 3.33 m/s NO! non-900 situation! 2.22 m/s current boat 270O

The boat is already due south (nothing needs to be done)
Rem? “Break each individual vector into its right angle components (NSEW).” 90O The boat is already due south (nothing needs to be done) 1800 0O 50.00 3.33 m/s 2.22 m/s 270O

Break the other into right angle components! South comp. West comp.
Function? Function? Sin θ = opp/hyp Cos θ = adj/hyp Sin = opp Cos = adj 2.22 m/s 2.22 m/s (1.43 m/s) 1800 0O 50.00 (1.70m/s) 3.33 m/s 2.22 m/s Cal= Cal= R= 1.43 m/s R= 1.70 m/s 270O

Consider the 3 remaining vectors.
Can any be added or subtracted? Added! The “Souths” can be________. (both are south) 5.03 m/s (1.43 m/s) 1800 0O 50.00 (1.70 m/s) = 1.70 3.33 m/s + m/s 3.33 m/s 270O

Note only two vectors remain…
They are at right angles! head tail Add them ______ to _______! 5.03 m/s (1.43 m/s) 1800 0O 50.00 (1.70 m/s) 3.33 m/s 270O

Now it’s a basic 2 vector problem. 5.03 m/s
magnitude A2 + B2 = C2 5.23 m/s C = A2 + B2 C = (1.43m/s)2 + (5.03 m/s)2 Cal= 1.43 m/s R= 5.23 m/s

As usual, choose the angle at the base of the resultant to be θ.
5.03 m/s θ direction Function? Tan θ = opp/adj Tan θ = 1.43 m/s/5.03 m/s Cal= 1.43 m/s R= 15.90 = 254.10

The concept of Equilibrium: CP and Honors
FYI… Often times in the real world there is a need to Prevent movement… Like when constructing a building- you want it to stay still. How is this done? 1st calculate the net force/movement in a situation… Then include an extra force that will perfectly negate the net force to keep it still. This force brings the structure back into equilibrium… so it is called the equilibrant force. clip..\..B

Equilibrant Force Problems
Two forces are on an object, one acts due east with 33.5 Newtons, the other acts due south with 45.5 N of force. What equilibrant force would need to be added to perfectly negate movement in this situation?

First without math… There are two forces on this object Let’s say that after working it out (tip to tail) we find this is the resultant force 33.5 N resultant 45.5 N

So this what happens overall…. What force could perfectly negate/cancel this force? 33.5 N resultant 45.5 N

One that is exactly the same size and in the opposite direction! (1800 away) This is the equilibrant force. 1800 1800 resultant 1800 away- so subtract or add 180) to find the angle

Tip to tail Now the MATH… 33.5 N 45.5 N 45.5 N

Now the MATH… 33.5 N 45.5 N

*Final answer must be from 00 (due east) Resultant magnitude:
Resultant direction: θ Tan θ = opp/adj Tan θ= 45.5/33.5 N 45.5 N Cal= (56.5 N) R= 53.60 *Final answer must be from 00 (due east) Resultant magnitude: A2 + B2 = C2 360 – 53.6= C = A2 + B2 C = N N2 R= 56.5 N Cal =

900 Net force = 0 (No Movement!) Equilibrant force 56.5 N 33.5 N 1800 00/3600 306.40 45.5 N 1800 56.5 N = 56.5 2700

new and improved-Vectors
now with… FRICTION!

Vectors Problems with Friction:
All the previous problems may also have a frictional force involved… We will revisit a few and see how friction will affect the answers.

Rope Pulling Problem w/ friction:
4.0 Kg 35.00 How much acceleration if µ= .333 ?

First treat it as if there is no friction… hyp 18.0N 4.00 Kg 35.00 ? (14.7 N) Which Function? Now what?... adj Cos θ = adj/hyp Cos = adj/18.0N Cross-multiply! Cal= R= 14.7 right (east)

Find the UP component of the pull hyp 18.0N opp (10.3N) 4.00 Kg 35.00 Which Function? Sin θ = opp/hyp Sin = opp/18.0N R= 10.3 N North (UP) Cal=

Rope Pulling Problems w/friction:
Find the weight.. (F=ma a=9.80ms/2 m=4.0 kg) 39.2 N (down) – 10.3N (up)= 28.9 N down So the actual weight is only 28.9N! hyp 39.2 N down 18.0N (10.3N) 4.00 Kg 35.00 * Note that the UP aspect is exactly opposite the weight 39.2N

Here’s the net weight of the box Box’s net weight (down) 39.2N -10.3N hyp (4.6 N) 28.9 N 18.0N 4.00 Kg 35.00 The ground must be pushing up with the same amount (normal force) 39.2N 28.9 N= FN

Here’s the net weight of the box Box’s net weight (down) 39.2N -10.3N 28.9 N 18.0N 4.00 Kg 35.00 µ= .333 (given) Ff= FN µ Ff = (28.9 N) (.333) 28.9 N= FN R= 9.62N of friction Cal=9.6237

Friction always opposes movement… Find the net force on the box… 14.7 N – 9.62 N = 5.1N to the right 9.62 N= Ff 18.0N 4.00 Kg 35.00 (14.7 N) R= 9.62 N of friction

Friction always opposes movement… Find the net force on the box… 14.7 N N = 5.1N to the right 9.62 N= Ff 18.0N 4.00 Kg 35.00 5.1 N= Fnet (14.7 N) R= 9.62 N of friction

Finally, find the acceleration… F=ma a= F/m 18.0N 4.00 Kg 35.00 5.1 N= Fnet Cal= 1.275 F= 5.1 N 5.1N R= 1.3 m/s2 to the right a= m= 4.00 kg 4.00 Kg a= ?

Incline Plane Problems w/ friction:
What is the acceleration as the crate slides down the incline? µ= .222 525 N 30.00

Cont… Down! In what direction is weight always? 525 N 30.00 525 N

Remember, there are two components to the weight...
Force perp. Cont… 30.00 525 N Remember, there are two components to the weight... The force perpendicular.

525 N Cont… 30.00 525 N Force parallel and the force parallel.

Which makes it slide down the incline?
Cont… 30.00 525 N Which makes it slide down the incline? The force parallel!

525 N Cont… 525 N The Ground angle and the angle θ shown
30.00 30.00 525 N The Ground angle and the angle θ shown will always be the same.

Which side from θ are we looking for? The Opposite side.
Cont… θ 30.00 30.00 525 N hyp opp Which side from θ are we looking for? The Opposite side. Which side is the Force of weight? The hypotenuse Which function? Sin θ = Opp/Hyp

525 N Cont… Mass of crate 525 N hyp Fw=ma Sin θ = Opp/Hyp m= F/a opp
(53.6 Kg) 525 N 263 N Cont… θ 30.00 30.00 Mass of crate 525 N hyp Fw=ma 263 N Sin θ = Opp/Hyp m= F/a opp Opp Sin = m= 525N/9.80 m/s2 525N 1 m= 53.6 kg Cal= 262.5 R= 263 N (II) down the incline if no friction

This counteracting force is the Force Normal
Note The force perpendicular is actually the force that presses the box against the incline (note its direction). The box does not fall through the incline because the incline is pushing back with exactly the same amount of force. (53.6 Kg) 525 N 263 N This counteracting force is the Force Normal θ FN 30.00 30.00 This pressing together of the block and inline causes friction…calculate it using (Ff = FN µ) 525 N hyp 263 N

1st Calculate the force perpendicular… 525 N
(53.6 Kg) 1st Calculate the force perpendicular… 525 N 263 N θ FN= 455N 30.00 30.00 adj 525 N 455N hyp Cos θ = Adj/Hyp Adj Cos = 525N 1 So the FN must be the same to counteract it. Cal= R= 455 N (Fperp)

Ff= FN µ Now calculate the Force of friction. (Ff) 525 N FN= 455N 455N
(53.6 Kg) 101 N Now calculate the Force of friction. (Ff) 525 N 263 N θ FN= 455N 30.00 30.00 455N µ= .222 (given) 525 N hyp Ff= FN µ This friction will fight the force down the incline. Ff = (455 N) (.222) Cal=101.01 R= 101 N of friction

Find the net force down the incline… 525 N
(53.6 Kg) 101 N Find the net force down the incline… 525 N 263 N θ 30.00 30.00 µ= .222 (given) 101 N 263 N 263N N= 162 N This is the net force down the incline.

Finally, Find the acceleration down the incline… 525 N
(53.6 Kg) Finally, Find the acceleration down the incline… 525 N 162 N θ 30.00 30.00 162 N = Fnet F=ma a= F/m Cal= F= 162 N 162N R= 3.02 m/s2 down the incline m= 53.6 kg 53.6 Kg a= ?

Let’s compare to the frictionless answer we got earlier in the notes….

What is the acceleration as the crate slides down the incline? F=ma
Cont… θ 30.00 30.00 525 N hyp opp What is the acceleration as the crate slides down the incline? F=ma a=F/m F=263 N (II) down the incline a= 263N/53.6 Kg Cal= m= 53.6 kg 4.91 m/s2

Bell work: date: 8.88 m/s2 to the right
A boy pulls a 5.00 kg sled with 49.0 N along the ground by a rope that has a ground angle of 25.0 degrees. What will the acceleration of the sled be along the ground? 8.88 m/s2 to the right

Summary of Basic Vector problems
Component vectors (practice finding angles- ex. ) Basic rope pulling frictionless & Friction Incline Plane frictionless & friction/.not sliding Hanging objects problems River Problems Equilibrant Force problems Non-900 problems (honors)

Bell work: date: Frictional force= 25N
A 5.0 kg box sits still on a incline plane. How much frictional force is present in this situation? *What is the coefficient of friction in this situation? Frictional force= 25N

Bell work: Date: A boat travels 6.60 m/s westward across a m river that flows 2.50 m/s southward. How far southward will the boat travel by the time it hits the opposite shore? What direction is it actually moving? 200.70 9.70 m

Vectors 2 Applying the Basics

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