Applications & Examples of Newton’s Laws

• Forces are VECTORS!!

• Newton’s 2nd Law: ∑F = ma

∑F = VECTOR SUM of all forces on mass m

Need VECTOR addition to add forces in the 2nd Law!– Forces add according to rules of VECTOR

ADDITION! (Ch. 3)

• Newton’s 2nd Law problems:

• STEP 1: Sketch the situation!!– Draw a “Free Body” diagram for EACH body in

problem & draw ALL forces acting on it.• Part of your grade on exam & quiz problems!

• STEP 2: Resolve the forces on each body into components– Use a convenient choice of x,y axes

• Use the rules for finding vector components from Ch. 3.

• STEP 3: Apply Newton’s 2nd Law to

EACH BODY SEPARATELY:

∑F = ma

– A SEPARATE equation like this for each body!– Resolved into components:

∑Fx = max ∑Fy = may

Notice that this is the LAST step, NOT the first!

Conceptual ExampleMoving at constant v, with NO friction,

which free body diagram is correct?

Example Particle in Equilibrium

“Equilibrium” ≡ The total force is zero. ∑F = 0 or ∑Fx = 0 & ∑Fy = 0

Example (a) Hanging lamp (massless chain).

(b) Free body diagram for lamp.∑Fy = 0 T – Fg = 0; T = Fg = mg

(c) Free body diagram for chain.∑Fy = 0 T – T´ = 0; T´ = T = mg

Example Particle Under a Net Force

Example (a) Crate being pulled to right across a floor.

(b) Free body diagram for crate.∑Fx = T = max ax = (T/m)

ay = 0, because of no vertical motion.

∑Fy = 0 n – Fg = 0; n = Fg = mg

Example Normal Force Again

“Normal Force” ≡ When a mass is in contact with a surface, the Normal Force n = force perpendicular to (normal to)

the surface acting on the mass. Example Book on a table. Hand pushing down.

Book free body diagram. ay = 0, because of no vertical motion (equilibrium).

∑Fy = 0 n – Fg - F = 0

n = Fg + F = mg + F Showing again that the normal force is not always = & opposite to the weight!!

Example 5.4: Traffic Light at Equilibrium

(a) Traffic Light, Fg = mg = 122 Nhangs from a cable, fastened to a support. Upper cables are weaker than vertical one. Will break if tension exceeds100 N. Does light fall or stay hanging? (b) Free body diagram for light. ay = 0, no vertical motion.

∑Fy = 0 T3 – Fg = 0 T3 = Fg = mg = 122 N(c) Free body diagram for cable junction (zero mass). T1x = -T1cos(37°), T1y = T1sin(37°) T2x = T2cos(53°), T2y = T2sin(53°), ax = ay = 0. Unknowns are T1 & T2.

∑Fx = 0 T1x + T2x = 0 or -T1cos(37°) + T2cos(53°) = 0 (1)∑Fy = 0 T1y + T2y – T3 = 0 or T1sin(37°) + T2sin(53°) – 122 N = 0 (2)

(1) & (2) are 2 equations, 2 unknowns. Algebra is required to solve for

T1 & T2! Solution: T1 = 73.4 N, T2 = 97.4 N

Example

Example 5.6: Runaway Car

Example 5.7: One Block Pushes Another

Example 5.8: Weighing a Fish in an Elevator

Example 5.9: Atwood Machine

Example 5.10Inclined Plane, 2 Connected Objects