Unit 2 Notes. Free Body Diagrams Show an object and the forces acting on it The object is represented by a circle (you can write the object’s name inside.

Unit 2 Notes

Free Body Diagrams

• Show an object and the forces acting on it

• The object is represented by a circle (you can write the object’s name inside the circle)

• Clearly labeled arrows pointing away from the object represent forces acting on the object – Their length represents the magnitude of the force

Good Example

Common Forces

• General push or pull• Gravity• Normal Force– 90 degrees from a surface

• Friction– Resists motion (always in the opposite direction from

motion)

For most forces, whatever exerts the force is listed first followed by the receiver of the force.

Components of Forces

When you are given a force and an angle, make a right triangle with the angle inside the triangle

SOHCAHTOA

is the measure of the angle

𝒔𝒊𝒏𝜽=𝒐𝒑𝒑𝒐𝒔𝒊𝒕𝒆

𝒉𝒚𝒑𝒐𝒕𝒆𝒏𝒖𝒔𝒆

𝒄𝒐𝒔 𝜽=𝒂𝒅𝒋𝒂𝒄𝒆𝒏𝒕

𝒉𝒚𝒑𝒐𝒕𝒆𝒏𝒖𝒔𝒆

𝒕𝒂𝒏𝜽=𝒐𝒑𝒑𝒐𝒔𝒊𝒕𝒆𝒂𝒅𝒋𝒂𝒄𝒆𝒏𝒕

Fopp = Fsin(ϴ)

Fadj = Fcos(ϴ)

Find the x- and y- components of the force

Example 2

Equations of Forces

∑Fy = Upward Forces – Downward Forces

∑Fx = Forces to Right – Forces to Left

(∑ means the summation)

For this unit,∑Fx = 0

∑Fy = 0

Equations of Forces

• When there is no acceleration, the sum of forces in the x-direction is 0 and the sum of forces in the y-direction is 0.

(∑ means the summation)

∑Fy = 0

∑Fx = 0

ExampleA person pushes a 10kg object to the left with a 20N force. Solve for all forces acting on the object.

Step1: Draw a free body diagram.

Step 2: Write the force summation equations.

∑Fx=0 ∑Fy=0

Step 3: Insert Forces into theequations. From here it is a 3-step solution.

Ff -FP/O=0 FN - Fg =0

Fg = 9.8N/kg * 10kg = 98N

Ff -FP/O=0 FN - Fg =0

Ff -20N=0 FN – 98N =0

Ff = 20N FN = 98N

• A person pulls on a 50kgdog’s leash. The dogdoesn’t move.

• Solve for all forces

Ff = 38.6N

∑Fx = 0

FLeash/Dog,x -Ff = 0

∑Fy = 0

38.6N -Ff = 0

FN + FLeash/Dog,y - Fg = 0

FN + 10.4N – 490N = 0

FN – 479.6N = 0

FN = 479.6N

Ramps

• Sometimes it is easier to redefine the x- and y-axes so that they are perpendicular and parallel to a surface

A 5kg block is at rest onan incline as shown.

Solve for the normal force and the force of friction.

Fg = 5kg * 9.8N/kg = 49N

Fg,x = 49N * sin(30) = 24.5N

Fg,y = 49N * cos(30) = 42.4N

-Ff + Fg,x = 0-Ff + 24.5N = 0

Ff = 24.5N

∑Fx = 0 ∑Fy = 0

FN - Fg,y = 0

FN – 42.4N = 0