Answer:
1(a) 55.0°
1(b) 58.3°
2(a) 10.2 N
2(b) 2.61 m/s²
3(a) 76.7°
3(b) 12.8 m/s
3(c) 3.41 s
3(d) 21.8 m/s
3(e) 18.5 m
4(a) 7.35 m/s²
4(b) 31.3 m/s²
4(c) 12.8 m/s²
Explanation:
1) Draw a free body diagram on the beam. There are five forces:
Weight force mg pulling down at the center of the beam,
Normal force Na pushing up at point A,
Friction force Na μa pushing left at point A,
Normal force Nb pushing perpendicular to the incline at point B,
Friction force Nb μb pushing up the incline at point B.
There are 3 unknown variables: Na, Nb, and θ. So we're going to need 3 equations.
Sum of forces in the x direction:
∑F = ma
-Na μa + Nb sin φ − Nb μb cos φ = 0
Nb (sin φ − μb cos φ) = Na μa
Nb / Na = μa / (sin φ − μb cos φ)
Sum of forces in the y direction:
∑F = ma
Na + Nb cos φ + Nb μb sin φ − mg = 0
Na = mg − Nb (cos φ + μb sin φ)
Sum of torques about point B:
∑τ = Iα
-mg (L/2) cos θ + Na L cos θ − Na μa L sin θ = 0
mg (L/2) cos θ = Na L cos θ − Na μa L sin θ
mg cos θ = 2 Na cos θ − 2 Na μa sin θ
mg = 2 Na − 2 Na μa tan θ
Substitute:
Na = 2 Na − 2 Na μa tan θ − Nb (cos φ + μb sin φ)
0 = Na − 2 Na μa tan θ − Nb (cos φ + μb sin φ)
Na (1 − 2 μa tan θ) = Nb (cos φ + μb sin φ)
1 − 2 μa tan θ = (Nb / Na) (cos φ + μb sin φ)
2 μa tan θ = 1 − (Nb / Na) (cos φ + μb sin φ)
Substitute again:
2 μa tan θ = 1 − [μa / (sin φ − μb cos φ)] (cos φ + μb sin φ)
tan θ = 1/(2 μa) − (cos φ + μb sin φ) / (2 sin φ − 2 μb cos φ)
a) If φ = 70°, then θ = 55.0°.
b) If φ = 90°, then θ = 58.3°.
2) Draw a free body diagram of each mass. For each mass, there are four forces. For mass A:
Weight force Ma g pulling down,
Normal force Na pushing perpendicular to the incline,
Friction force Na μa pushing parallel down the incline,
Tension force T pulling parallel up the incline.
For mass B:
Weight force Mb g pulling down,
Normal force Nb pushing perpendicular to the incline,
Friction force Nb μb pushing parallel up the incline,
Tension force T pulling up the incline.
There are four unknown variables: Na, Nb, T, and a. So we'll need four equations.
Sum of forces on A in the perpendicular direction:
∑F = ma
Na − Ma g cos θ = 0
Na = Ma g cos θ
Sum of forces on A up the incline:
∑F = ma
T − Na μa − Ma g sin θ = Ma a
T − Ma g cos θ μa − Ma g sin θ = Ma a
Sum of forces on B in the perpendicular direction:
∑F = ma
Nb − Mb g cos φ = 0
Nb = Mb g cos φ
Sum of forces on B down the incline:
∑F = ma
-T − Nb μb + Mb g sin φ = Mb a
-T − Mb g cos φ μb + Mb g sin φ = Mb a
Add together to eliminate T:
-Ma g cos θ μa − Ma g sin θ − Mb g cos φ μb + Mb g sin φ = Ma a + Mb a
g (-Ma (cos θ μa + sin θ) − Mb (cos φ μb − sin φ)) = (Ma + Mb) a
a = -g (Ma (cos θ μa + sin θ) + Mb (cos φ μb − sin φ)) / (Ma + Mb)
a = 2.61 m/s²
Plug into either equation to find T.
T = 10.2 N
3i) Given:
Δx = 3.7 m
vᵧ = 0 m/s
aₓ = 0 m/s²
aᵧ = -10 m/s²
t = 1.25 s
Find: v₀ₓ, v₀ᵧ
Δx = v₀ₓ t + ½ aₓ t²
3.7 m = v₀ₓ (1.25 s) + ½ (0 m/s²) (1.25 s)²
v₀ₓ = 2.96 m/s
vᵧ = aᵧt + v₀ᵧ
0 m/s = (-10 m/s²) (1.25 s) + v₀ᵧ
v₀ᵧ = 12.5 m/s
a) tan θ = v₀ᵧ / v₀ₓ
θ = 76.7°
b) v₀² = v₀ₓ² + v₀ᵧ²
v₀ = 12.8 m/s
3ii) Given:
Δx = D cos 57°
Δy = -D sin 57°
v₀ₓ = 2.96 m/s
v₀ᵧ = 12.5 m/s
aₓ = 0 m/s²
aᵧ = -10 m/s²
c) Find t
Δx = v₀ₓ t + ½ aₓ t²
D cos 57° = (2.96 m/s) t + ½ (0 m/s²) t²
D cos 57° = 2.96t
Δy = v₀ᵧ t + ½ aᵧ t²
-D sin 57° = (12.5 m/s) t + ½ (-10 m/s²) t²
-D sin 57° = 12.5t − 5t²
Divide:
-tan 57° = (12.5t − 5t²) / 2.96t
-4.558t = 12.5t − 5t²
0 = 17.058t − 5t²
t = 3.41 s
d) Find v
vₓ = aₓt + v₀ₓ
vₓ = (0 m/s²) (3.41 s) + 2.96 m/s
vₓ = 2.96 m/s
vᵧ = aᵧt + v₀ᵧ
vᵧ = (-10 m/s²) (3.41 s) + 12.5 m/s
vᵧ = -21.6 m/s
v² = vₓ² + vᵧ²
v = 21.8 m/s
e) Find D.
D cos 57° = 2.96t
D = 18.5 m
4) Given:
R = 90 m
d = 140 m
v₀ = 0 m/s
at = 0.7t m/s²
The distance to the opposite side of the curve is:
140 m + (90 m) (π/2) = 281 m
a) Find Δx and v if t = 10.5 s.
at = 0.7t
Integrate:
vt = 0.35t² + v₀
vt = 0.35 (10.5)²
vt = 38.6 m/s
Integrate again:
Δx = 0.1167 t³ + v₀ t + x₀
Δx = 0.1167 (10.5)³
Δx = 135 m
The car has not yet reached the curve, so the acceleration is purely tangential.
at = 0.7 (10.5)
at = 7.35 m/s²
b) Find Δx and v if t = 12.2 s.
at = 0.7t
Integrate:
vt = 0.35t² + v₀
vt = 0.35 (12.2)²
vt = 52.1 m/s
Integrate again:
Δx = 0.1167 t³ + v₀ t + x₀
Δx = 0.1167 (12.2)³
Δx = 212 m
The car is in the curve, so it has both tangential and centripetal accelerations.
at = 0.7 (12.2)
at = 8.54 m/s²
ac = v² / r
ac = (52.1 m/s)² / (90 m)
ac = 30.2 m/s²
a² = at² + ac²
a = 31.3 m/s²
c) Given:
Δx = 187 m
v₀ = 0 m/s
at = 3 m/s²
Find: v
v² = v₀² + 2aΔx
v² = (0 m/s)² + 2 (3 m/s²) (187 m)
v = 33.5 m/s
ac = v² / r
ac = (33.5 m/s)² / 90 m
ac = 12.5 m/s²
a² = at² + ac²
a = 12.8 m/s²
