Answer:
a = 3.27 m/s²
F = 32.7 N
Explanation:
Draw a free body diagram. There are three forces:
Weight force mg pulling straight down.
Normal force N pushing perpendicular to the slope.
Friction force F pushing parallel up the slope.
Sum of forces in the parallel direction:
∑F = ma
mg sin θ − F = ma
Sum of torques about the cylinder's axis:
∑τ = Iα
Fr = ½ mr²α
F = ½ mrα
Since the cylinder rolls without slipping, a = αr. Substituting:
F = ½ ma
Two equations, two unknowns (a and F). Substituting the second equation into the first:
mg sin θ − ½ ma = ma
Multiply both sides by 2/m:
2g sin θ − a = 2a
Solve for a:
2g sin θ = 3a
a = ⅔ g sin θ
a = ⅔ (9.8 m/s²) (sin 30°)
a = 3.27 m/s²
Solving for F:
F = ½ ma
F = ½ (20 kg) (3.27 m/s²)
F = 32.7 N
