1. Frictionless incline. Force F F F m = 1 0 k g m=10kg m = 1 0 k g F F F m m m F N F_N F N
Solution
a. Constant velocity indicates zero acceleration.
∑ F x = F x − W x = 0 F x = F cos θ W x = m g sin θ ⇒ F = m g sin θ cos θ = 1 0 ∗ 9 . 8 ∗ sin 3 0 ∘ cos 3 0 ∘ ≈ 5 7 . 7 N \begin{aligned}
\sum F_x &= F_x - W_x = 0\\
F_x &= F\cos \th\\
W_x &= mg\sin \th\\
\To F &= \frac{mg \sin \th}{\cos\th} = \frac{10*9.8*\sin 30^{\circ}}{\cos 30^{\circ}} \approx 57.7 N
\end{aligned}
∑ F x F x W x ⇒ F = F x − W x = 0 = F cos θ = m g sin θ = cos θ m g sin θ = cos 3 0 ∘ 1 0 ∗ 9 . 8 ∗ sin 3 0 ∘ ≈ 5 7 . 7 N
b. In y y y
∑ F y = F N − W y − F y = 0 W y = m g cos θ F y = F sin θ ⇒ F N = m g cos θ + F sin θ = 1 0 ∗ 9 . 8 ∗ cos 3 0 ∘ + 5 7 . 7 sin 3 0 ∘ ≈ 1 1 5 . 5 N \begin{aligned}
\sum F_y &= F_N - W_y - F_y = 0\\
W_y &= mg\cos \th\\
F_y &= F\sin\th\\
\To F_N &= mg\cos \th + F\sin \th\\
& = 10*9.8*\cos 30^{\circ} + 57.7 \sin 30^{\circ} \approx 115.5N
\end{aligned}
∑ F y W y F y ⇒ F N = F N − W y − F y = 0 = m g cos θ = F sin θ = m g cos θ + F sin θ = 1 0 ∗ 9 . 8 ∗ cos 3 0 ∘ + 5 7 . 7 sin 3 0 ∘ ≈ 1 1 5 . 5 N
2. Find maximum velocity v v v
Solution
Normal force F N F_N F N F s = F N ⋅ u s F_s = F_N \cdot u_s F s = F N ⋅ u s F c = m v 2 R F_c = \dfrac{mv^2}{R} F c = R m v 2
m g = F N cos θ − F s sin t h = F N ( cos θ − u k sin t h ) ⇒ F N = m g cos θ − u k sin t h Eq 1 F c = F N sin θ + F s cos θ = F N ( sin θ + u k cos θ ) = m g cos θ − u k sin θ ⋅ ( sin θ + u k cos θ ) F c = m v 2 R ⇒ v = g R ( sin θ + u k cos θ ) cos θ − u k sin θ = 9 . 8 ⋅ 1 0 0 ⋅ ( sin 3 0 ° + 0 . 4 ⋅ cos 3 0 ° ) cos 3 0 ° − 0 . 4 ⋅ sin 3 0 ° ≈ 3 5 . 2 9 m / s \begin{aligned}
mg &= F_N \cos \th - F_s\sin th\\
&= F_N (\cos \th - u_k\sin th)\\
\To & F_N=\frac{mg}{\cos \th - u_k\sin th} &\text{Eq 1}\\
F_c &= F_N \sin \th + F_s \cos \th\\
&= F_N (\sin \th + u_k \cos \th) \\
&= \frac{mg}{\cos \th - u_k\sin \th} \cdot (\sin \th + u_k \cos \th)\\
F_c &= \frac{mv^2}{R}\\
\To &v = \sqrt {\frac{gR(\sin \th + u_k \cos \th)}{\cos \th - u_k\sin \th}}\\
&= \sqrt {\frac{9.8\cdot 100 \cdot (\sin 30^\degree + 0.4 \cdot \cos 30^\degree)}{\cos 30^\degree - 0.4\cdot \sin 30^\degree}} \approx 35.29m/s
\end{aligned}
m g ⇒ F c F c ⇒ = F N cos θ − F s sin t h = F N ( cos θ − u k sin t h ) F N = cos θ − u k sin t h m g = F N sin θ + F s cos θ = F N ( sin θ + u k cos θ ) = cos θ − u k sin θ m g ⋅ ( sin θ + u k cos θ ) = R m v 2 v = cos θ − u k sin θ g R ( sin θ + u k cos θ ) = cos 3 0 ° − 0 . 4 ⋅ sin 3 0 ° 9 . 8 ⋅ 1 0 0 ⋅ ( sin 3 0 ° + 0 . 4 ⋅ cos 3 0 ° ) ≈ 3 5 . 2 9 m / s Eq 1
