Physics Notes, Unit 4, Free Particle

Force Diagrams

Force Diagrams

•Fkind of force, on feeler, by dealer (for example: Fgravity, on student, by earth)

• System: the object under consideration as the

• Surroundings: everything else in the environment that might in

any significant way affect the system.

• Fkind of force, on feeler, by dealer (for example: Fgravity, on

student, by earth)

Force Diagrams •In general, we will follow the following steps when creating force diagrams. 1. Sketch the system and its surroundings. 2. Enclose the system within a system boundary. 3. Shrink the system to a point at the center of coordinate axes with one axis parallel to the direction of motion. 4. Represent all relevant forces (across the system boundary) with a

labeled vector.

5. Force vectors usually point away from object

6. Indicate which forces (if any) are equal in magnitude to

other forces.

Since the shape of the object is

unimportant, shrink it to a point.

Place it at the intersection of a set

of coordinate axes with one of the

axes parallel to the direction of

motion as shown in figure 4.

Fkind of force, on feeler, by

dealer

Fnet

0

Fkind of force, on feeler, by

dealer

Section

4.1 Force and Motion

A contact force exists when

an object from the external

world touches a system

and thereby exerts a force

on it.

Contact Forces and Field Forces

Section 4.1

Fnet

Fnet

0

Fkind of force, on feeler, by

dealer

A car parked on a hill

A car parked on a hill

• Gravity always points

toward the center of the

earth (down).

• The normal force is

perpendicular to the

road/tire surface and the

friction force is parallel to

the road/tire surface.

• Friction exerts a force up

the hill to resist the

tendency of the car to

slide down the hill due to

gravity.

A car parked on a hill

• Gravity always points

toward the center of the

earth (down).

• The normal force is

perpendicular to the

road/tire surface and the

friction force is parallel to

the road/tire surface.

• Friction exerts a force up

the hill to resist the

tendency of the car to

slide down the hill due to

gravity.

• Since the car is motionless, the forces must be balanced along each coordinate axis.

• Gravity isn't along either coordinate axis, but we can represent gravity with two component vectors.

• Fg parallel is how much of the gravitational force tends to pull the car along the slope

• and Fg perpendicular is how much of the gravitational force tends to pull the car to the road.

• For the forces to be balanced, Fg parallel must be equal in size to Ffriction and Fg perpendicular must be equal in size to Fnormal

A car parked on a hill

FN

Fg

Ff

Fpush

Fpush y (downward)

component

Fpush x (horizontal) component

These two diagrams

show the same thing

even though we moved

the upward component

of tension.

Tension force with x & y components

All forces are balanced

Physics Bell Work, Wednesday, Nov 5 4. You accelerate a grocery cart along a level floor in the presence

of friction effects between the cart and the floor. Draw force diagrams for you, and the cart. Fully label all vectors. Show the force pairs.

*

□

Fnet

you

FN, y, f

=

=

Fg, y,

FN, y, c Ffs, y, f

you

Ff, y, f

e

Floor/ earth

There are 7 Third Law pairs in the diagrams. All pairs have the same symbol next to them

If The cart is speeding up, how does the size of the frictional force

on the cart by the floor compare to the frictional force on you by the floor? The frictional force on you must be larger.

Problems with Trig

3. Find the horizontal and vertical components of the tension in the fishing line. Draw the tension force on the fish and include the x & y components of the tension.

Objective: break forces not aligned with your coordinate axis into components using trigonometry.

Tension

y

Ty

𝑨 • 𝐬𝐢𝐧𝜽 = 𝑨𝒚

𝑨 • 𝐜𝐨𝐬𝜽 = 𝑨𝒙

Problems with Trig

1. Find the horizontal and vertical components of the tension in the fishing line. Draw a force diagram that include the components.

Objective: break forces not aligned with your coordinate axis into components using trigonometry.

𝟑𝟖 𝑵 • 𝐜𝐨𝐬 𝟓𝟓° = 𝑻𝒙 = 22 N

yToppsinθ*hyp

𝟑𝟖 𝑵 • 𝐬𝐢𝐧 𝟓𝟓° = 𝑻𝒚 = 31 N

= Tx

cos adj

hyp

hyp

oppsin

Adjacent:

Opposite:

Problems with Trig

Fg y or perpendicular Fg x or parallel

2. A 2000 kg elephant stands on a ramp. Determine the components of the elephant’s weight parallel and perpendicular to the ramp.

𝜽

𝜽

𝜽= 20°

= opposite

= adjacent

Adjacent:

cos adj

hyp

𝑨dj = perpendicular (Fg −

y) = 20,000 N• cos 20° = 18794 N

Fg Elephant, Earth

Fg = w = mg = 2000kg ⦁ 10 N/kg = 20,000 N

Opposite:

𝑶𝒑𝒑= parallel (Fg −

x) = 20,000 N• sin 20° = 6840 N

hyp

oppsin yToppsinθ*hyp

= 20,000 N

x

y

Problems with Trig

1. A 950 kg car is driven up a hill at constant velocity of 7 m/s, where 1200 N of friction and drag oppose its motion. I. Draw a force diagram for the car. II. What is the weight of the car? Fg = mg = 950kg (10 N/kg ) = 9500N 2. Is the normal force on the car a. > than the weight b. < the weight c. = weight Because a2 + b2 = c2, weight = c2 3. How would you calculate the traction force on the car that allows it to go up the hill? (Ff) + Fg-x = Ft ||

F t

= 9500N

Fg-x, ||

Fg-y

= weight

Problems with Trig

1. A 950 kg car is driven up a hill at constant velocity of 7 m/s, where 1200 N of friction and drag oppose its motion. a. Draw a force diagram for the car. b. What is the weight of the car? Fg = mg = 950kg(10 ) = 9500N 4. Calculate the normal force on the car. FN = Fg-y Fg-y = Fg •cos22° FN = Fg-y (9500N) cos22° = 8800N 5. Calculate the force on the car that allows it to go up the hill. (Ffs) + Fg-x = Ft ||

Fg-x = Fg • sin 22°

(1200N) + (9500N) sin22° = Ftraction

1200N + 3600N = 4800N = Ftraction

Ft

= 9500N

Fg-x

Fg-y x

y