Thurs 24 Sep (Two-Body Tension)

AP Physics: Thurs SEP 24:

Target:

Learning Objective: Free Body Diagram

Journal:

5000m

10⁰

A 100,000 kg train climbs a 5,000 m mountain track at a speed of 20m/s at an angle of 10˚ with respect to the horizontal.

1. How much altitude does it gain?

a. 868 m

b. 1230 m

c. 3410 m

d. 4920 m

2. How much work did the train do?

3. How much power did the train use in the climb?

1. a (sin 10⁰)(5000) = 868m

2. W = (force)(dist)

or (weight) height) when lifting up a vertical

W = mg (height)

(100,000 x 10)(868) W = 868,000,000J

3. Power = W/time

time = dist/speed = 5000/20

= 250s

868,000,000J / 250s

Power = 3,472,000 W

Notes

Force: Push or Pull between TWO objects:

Contact Forces:

Normal - (Push always & 90⁰ from surface

sometimes called Applied (or normal) if not from below                               (Push always & 90⁰ from surface)

Tension - (Pull always)

Buoyancy (push up)

Friction (push against)

Force of Air or Drag (Air Friction)

Non-contact Forces

Gravitational Force (Pull always called Weight = Mass x Gravity)

Magnetism (push or pull)

Electric (push or pull)

Free Body Diagram: Means one object free from all other things except the forces on it. ( i.e., show only the forces acting on the object)

A book is at rest on a table top. A free-body diagram for this situation looks like this:

2. A girl is suspended motionless from the ceiling by two ropes. A free-body diagram for this situation looks like this:

3. An egg is free-falling from a nest in a tree. Neglect air resistance. A free-body diagram for this situation looks like this:

4. A flying squirrel is gliding (no wing flaps) from a tree to the ground at constant velocity. Consider air resistance. A free-body diagram for this situation looks like this:

5. A rightward force is applied to a book in order to move it across a desk with a rightward acceleration. Consider frictional forces. Neglect air resistance. A free-body diagram for this situation looks like this:

6. A rightward force is applied to a book in order to move it across a desk at constant velocity. Consider frictional forces. Neglect air resistance. A free-body diagram for this situation looks like this:

7. A college student rests a backpack upon his shoulder. The pack is suspended motionless by one strap from one shoulder. A free-body diagram for this situation looks like this:

8. A skydiver is descending with a constant velocity. Consider air resistance. A free-body diagram for this situation looks like this:

9. A force is applied to the right to drag a sled across loosely-packed snow with a rightward acceleration. A free-body diagram for this situation looks like this:

10. A football is moving upwards towards its peak after having been booted by the punter. A free-body diagram for this situation looks like this:

11. A car is coasting to the right and slowing down. A free-body diagram for this situation looks like this:



No friction, what does the free body diagram look like?

Balanced forces:

Screenshot 03

Block moving at constant velocity.

Screenshot 05

Unbalanced force:

In the statement of Newton's first law, the unbalanced force refers to that force which does not become completely balanced (or canceled) by the other individual forces. If either all the vertical forces (up and down) do not cancel each other and/or all horizontal forces do not cancel each other, then an unbalanced force exists. The existence of an unbalanced force for a given situation can be quickly realized by looking at the free-body diagram for that situation. Free-body diagrams for three situations are shown below. What is the acceleration in each case?

What is the Net Force in each of the following cases?

Acceleration of 4 m/s²

Acceleration of - 4 m/s²

Newton's second law of motion pertains to the behavior of objects for which all existing forces are not balanced. The second law states that the acceleration of an object is dependent upon two variables - 1. net force acting upon the object, and

2. inversely upon the mass of the object.

Journal:

1. A lump of soft clay of mass m falls from a height h above the floor, where it crashes and comes to rest. What force brings the clay to a stop? How much work is done by the floor on the clay? What is the net amount of work done on the clay during the fall? Where does that energy end up?

Key: The floor pushing up on the clay stops it. Only insofar as the floor distorts by some amount Δy will there be any work done on the clay (W = F times Δy). That's because the point of application of the force doesn't appreciably move (Δy = 0). The gravitational field does work on the clay equal to W = mgh. The clay distorts its atoms more with respect to each other and that impart random KE (i.e., thermal energy) to the clay—it gets warmer. The energy mgh which is external becomes internal via the interatomic interactions as the clay distorts and comes to rest.

Chapter 4:

(forces)

1. T₁ Find: T₁ =

40kg T₂ =

T₂

50kg

1. T₁= 900N, T₂ = 500N

2. A 40K g object hangs from two ropes as shown:

What is the Tension ( T₁& T₂) in the two ropes?

37⁰ 53⁰

T₁T₂

40kg

(X): (Cos 37) T₁ = (Cos 53) T₂

T₁ = .602/.800 T₂ à .753 T₂

(Y): 400N = (Sin 37) T₁ + (Sin 53) T₂

400N = .602 (.753 T₂) + .800 T₂

400 = .453 T₂+ .800 T₂ à 1.25 T₂

T₂ = 400/ 1.25 = 320N

T₁ = .753 T₂ à .753(320) = 241N

*Free-body diagrams:

N (or F_N)

F_KN or (F_N)

mg (or W)

3. T₃T₂T₁ =18N

M₃ m₂ m₁

d3. Three massive blocks of m₁ = 10Kg, m₂ = 15kg, and m₃ = 20kg. They are pulled on a frictionless table by a force of 18N. What are the tensions in the T₂ and T₃ ropes?

F_net = ma à 18 = 45a à .4 m/s²

a) T₂: F_net(2) = m₂a₂ à (35kg)( .4 N/kg) = 14N

b) T₃: F_net(3) = m₃a₃ à (20kg)( .4 N/kg) = 8N

4. A small marble rolls across floor at 1.50m/s and off the top of a flight of stairs. Each step is 20cm high and 20cm wide. On what step (from the top) will the marble hit?

a. _____

s_y = ½ at² à t = √ (.2/5) = .2 sec

s_x = vt à (1.5)(.2) = .3m (farther than 1^st step .2m )

b. _____

s_y = ½ at² à t = √ (.4/5) = .28 sec

s_x = vt à (1.5)(.28) = .42m (farther than 2^nd step .4m)

b. _____

s_y = ½ at² à t = √ (.6/5) = .35 sec

s_x = vt à (1.5)(.35) = .52m (on 3^rd step .6m)

5. A 60kg man is able to toss a 12kg object 2.0 m in the air directly above him. The toss takes 0.30 sec. Find: a. What is velocity of the object at the start of the toss? b. What was the impulse?

c. What is the acceleration of the object at the apex? d. What is the force on the ground just as the man releases the object?

a. PE = KE à mgh = ½ mv² à

V = √ (10)(2)(2) = 6.32m/s

b. mDV = fDt à (12)(6.32) = 75.9 kg m/s

c. g = – 10ms²

d. Dmomentum = impulse à 75.9 = fDt à 75.9 = f(.3) à f = 253N

6. A man shoots a gun at an apple 25m high in a tree (54.0m away). Just as he shoots, the apple falls from the branch (next to the tree). a. How close does the bullet get to the falling apple (No air friction)? b. The bullets velocity is 150m/s. How long will the bullet take to hit the tree? c. How far will the apple fall (as bullet hits the tree)?

a. Hits the apple. Both fall at the same rate

b. S_x = vt à 54 = 150t à t = .36s

c. S_y = V_ot + ½ at²

S_y = 0 + (.5)(-10)(.36²) = .648m

7. What is the

a. Mechanical Advantage

b. Effort Force (Resistance force is 100N)

c. Effort Distance (Resistance dist. is 2.0m)

I. incline plane of 30⁰

II. Wheel (2.4m) & Axel (0.2m)

III. Pulley of 1 weight bearing line

III. Pulley of 3 weight bearing lines

IV. Pulley of 4 weight bearing lines

V. 12m Lever: 2m on resistance side of fulcrum and 10m on effort side

VI. 16m incline plane that is 4m high

I. a. Sin 30⁰ = .5, so MA = 2

b. 50N (½ the force)

c. 4.0 m (twice the resistance distance)

II. a. 2.4/0.2 = 12, so MA = 12

b. 8.33N (100/12)

c. 24m (2m)(12)

III. a. MA = 1

b. 100N

c. 2m

IV. a. MA = 4

b. 25N (100/4)

c. 8m (2m)(4)

V. a. MA = 5

b. 20N (100/5)

c. 10m (2m)(5)

VI. a. MA = 4

b. 25N (100/4)

c. 8m (2m)(4)

8. A 170kg hot air balloon hovers at a constant altitude above the ground. One 70kg person is in the gondola while a 80kg person is hanging from a rope below the gondola. The person on the rope starts to ascend at constant rate at 1.0m/s.

a. What is the direction and speed of the balloon as he ascends?

b. If the person climbs for 10 seconds and then stops, what is the change in the balloons altitude?

a. Momentum (Mo) is conserved

Mo up = Mo down

mV = mV à (80)(1) = (240)(V)

V = 0.333 m/s down

b. Person below went up 10m: (1) (10)

PE is conserved (no KE)

mgh = mgh

(80)(10)(10) = (240)(10)(y)

y = 3.33m down

vt = (.333)(10) = 3.33m

9. A trunk is pulled across a floor at a constant speed by a 200N force applied by a rope at an angle of 60⁰ above the horizontal. If the trunk weighs 150N. The coefficient of friction is:

F_K = F_horz

(m) F_N = (Cos 60)(200)

(m) (mg - Sin 60 mg) = (Cos 60)(200)

(m) (150 - 130) = (.5)(200)

m = 100 / 20 = 5

10. A man pushes a 10kg baby up a 30⁰ incline at a constant speed. He applies 60N force parallel to the slope. a. What is the force of Kinetic friction? b. What is kinetic m?

F_N = (Cos q)(mg)

F_x = (Sin q)(mg)

F_x = (Sin q)(mg) = (.5)(10)(10) =50N (no friction

60N – 50N = 10N of Kinetic friction

F_K = mF_N = m (Cos 30)(mg)

à 10 = m (.866)(10)(10)

m = .115

11. The block is resting on an incline plane (static m = 0.30 and kinetic m = 0.18). The angle is increased until the block starts to move.

a. What is the angle (block starts to move)?

b. What is the block’s acceleration at this angle?

c. What is the speed of the block when it hits the ground (the length of the incline plane = 4.0m)?

d. After reaching the floor, the block slides with the same kinetic m (0.18). How far does it slide?

F_x = (Sin q)(mg) , F_S = m_SF_N , F_N = (Cos q)(mg)

F_x = F_S à (Sin q)(mg) = m_S(Cos q)(mg)

à (Sin q) = m_S(Cos q) à m_S= (Sin q)/(Cos q)

à Tanq = m_S à Tanq = .3 à Tanq ^-1 (.3) = 16.7⁰

F_x = (Sin q)(mg) , F_K = m_KF_N , F_N = (Cos q)(mg)

F_net = ma à F_x – F_K = ma

à (Sin q)(mg) – m_K(Cos q)(mg)= ma

à (Sin 16.7)(g) – 0.18(Cos 16.7)(g)= a

à (.287)(10) – 0.18(.958)(10)= a

à (2.87) – (1.72) = a à a = 1.15m/s²

V_f² – V_o² = 2aDs

V_f² – 0 = 2(1.15)(4)

V_f = √ 9.2 = 3.03m/s

F_net = ma à F_K = ma

m_K(mg) = ma à m_K(g) = a

a = 0.18(10) = -1.8m/s²

V_f² – V_o² = 2aDs

0 – (3.03)² = 2 (-1.8)s à s = 5.6m

a T =

4kg

5kg

12. A 5kg block hangs over a table and pulls a 4kg block on the table with m_K = 0.4. What is the:

a. Net force?

b. acceleration?

c. tension in line?

F_net= W₅ - F_K

F_net=m₅g – (mm₄g)

F_net= 50 - (.4)(40) à 16N

F_net= 50 - 16N = 34N

F_net= m_totala

34 = 9a à a = 3.8m/s²

F_net= m₅a

W – T = m₅a

50 –T = (5)(3.8)

T = 50 - 18.9 à 31.1N

13. A 10.0m board (100kg) is hanging 2.5m over the edge of a platform (that is above the water). Jimmy wants to take a good look at the water. If Jimmy (120kg) starts walking towards the end of the board, will he make it all the way without falling in? If not, how far can he walk on the board before both he and the board fall into the water below?

Jimmy M.

Cloud Callout: I’m frightened

7.5m 2.5m

GOC of board is 5m à 2.5m from edge

Torque from platform edge:

fd (ccw) = fd (cw)

(1000)(2.5) = 1200(x)

x = 2.08m

Yes, sorry to say, Jimmy would fall!