Unit 4 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

Common Forces

• General push or pull– The name of whatever is exerting

the force / name of what the force is exerted upon

• Gravity– Always pulls straight down

• Normal Force– 90 degrees from a surface

• Friction– Resists motion (always in the

opposite direction from where the object is trying to move)

• Tension– Pulls in the direction of the cable

NOTE: If you can’t tell me what is exerting the force, then that force probably doesn’t exist.

• Show all the forces on the block in the picture.

Solution:1) There is gravity.2) Perpendicular to the surface is

straight upward.3) There’s no reason to think it’s

frictionless. The block wants to move left, so friction is to the right.

4) There is a force F pushing on the block.

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. The object does not move. Solve for all forces acting on the object.

Step1: Draw a free body diagram.

Step 2: Write the force summation equations.

∑Fx= Ff +FP/O=0 ∑Fy=FN + Fg =0

Step 3 : Solve for the gravityand any components.

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

Step 4: Substitute numbers into the equations and solve.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

∑Fx = 0

FLeash/Dog,x +Ff = 0

∑Fy = 0

FN + FLeash/Dog,y + Fg = 0

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

sin(ϴ) = (Fopp)/Fcos(ϴ) = (Fadj)/F

Find the x- and y- components of the force

Example 2

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

• Solve for all forces

∑Fx = 0

FLeash/Dog,x +Ff = 0

∑Fy = 0

FN + FLeash/Dog,y + Fg = 0

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

Inclines

• 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 = mgFg = 5kg * 9.8N/kg = 49N

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

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

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

∑Fx = 0 ∑Fy = 0

FN + Fg,y = 0

FN – 42.4N = 0

Another way of looking at it.

• If it helps you visualize which parts of which forces are on which axis, you can rotate the free body diagram.

• Remember that the original is your free body diagram and the rotated diagram is not anything you need to do.