P Lab Friction

Define static and kinetic friction and when each applies.

• Static friction prevents motion; acts when object is at rest.

• Kinetic friction acts when the object is moving.

Why does increasing the angle of an inclined plane eventually cause an object to slide?

As angle increases, gravity’s component along the incline increases → overcomes static friction.

What equation gives the coefficient of static friction based on incline angle?

μs​=tanθ

Does adding mass to an object affect the coefficient of friction? Explain.

No. μ depends on materials, not mass.
Extra mass increases both friction and normal force equally.

What is the direction of the frictional force on an inclined plane?

Friction always opposes motion → up the incline if object is sliding down.

A 2.5 kg box is placed on an incline. It starts sliding at θ=25. Find the coefficient of static friction.

μs​=tan(25∘)≈0.466

A 0.1 kg sled is pulled by a 0.15 kg hanging mass. Distance = 0.5 m. Time measured: t=1.2 

a) Calculate acceleration
b) Total system mass
c) Net force
d) Find kinetic friction force
e) If normal force = 2.5 N, calculate μₖ

• m=0.1+0.15=0.25kg

• d=0.5 m,t=1.2 sd = 0.5 \, \text{m}, t = 1.2 \, \text{s}d=0.5m,t=1.2s

a) a= ( 2 × 0.5 )/1.2² = 0.694 m/s²

b) m=0.25 kg

c) F=0.15⋅9.8=1.47 N

d) fk = F -ma =1.296 N

e)
If N=2.5N = 2.5N=2.5, then: uk= fk/N = 0.518

You conducted a trial with:

• Hanging mass = 2.0 N → pulling force F

• Sled + bar = 0.6 kg → normal force N=5.9 

• Distance = 0.5 m

• Time t=1.0 s

8. Calculate:

a) a=2d/t²
b) m⋅a
c) fk=F−ma
d) Compare to μk⋅N with μk​=0.2

a) a= (2 × 0.5)/ 1² = 1m/s²

b)
Total mass: F=2N = m hang x g → m hang = 2/9.8 = 0.204

c)

fk=F−ma=2.0−0.804=1.196 Nf_k = F - ma = 2.0 - 0.804 = 1.196 \, \text{N}fk​=F−ma=2.0−0.804=1.196N

d)

μk⋅N=0.2⋅5.9=1.18 N\mu_k \cdot N = 0.2 \cdot 5.9 = 1.18 \, \text{N}μk​⋅N=0.2⋅5.9=1.18N

✔ Very close → confirms kinetic friction model.