Answer :
Answer:
vmin = 2.816 m/s
vmax = 3.1517 m/s
Explanation:
We apply Newton's Second Law as follows
∑Fy = m*g ⇒ TAC*Cos 30° + TBC*Cos 45° = m*g (1)
∑Fx = m*v²/R ⇒ TAC*Sin 30° + TBC*Sin 45° = m*v²/R (2)
Subtract Eq. (1) from Eq. (2). Since Sin 45° = Cos 45°, we obtain
TAC*(Cos 30°-Sin 30°) = (m*g) - (m*v²/R) (3)
Multiply Eq. (1) by Sin 30°, Eq. (2) by Cos 30°, and subtract:
TBC*(Sin 30°*Cos 45°-Cos 30°*Sin 45°) = Sin 30°*m*g - Cos 30°*(m*v²/R)
- TBC*Sin 15° = Sin 30°*m*g - Cos 30°*(m*v²/R) (4)
Making
TAC = 35 N, m = 4 Kg, R = 1.2 m, g = 9.81 m/s² in Eq. (3), we find the value v₁ of v for which TAC = 35 N:
35 N*(Cos 30°-Sin 30°) = (4 Kg*9.81 m/s²) - (4 Kg*v₁²/1.2 m)
⇒ 12.8109 = 39.24 - 3.33*v₁² ⇒ v₁ = 2.816 m/s
We have TAC ≤ 35 N for v ≥ v₁, that is, for v ≥ 2.816 m/s
Making
TBC = 35 N, m = 4 Kg, R = 1.2 m, g = 9.81 m/s² in Eq. (4), we find the value v₂ of v for which TBC = 35 N:
- 35 N*Sin 15° = Sin 30°*4 Kg*9.81 m/s² - Cos 30°*(4 Kg*v₂²/1.2 m)
⇒ - 9.059 N = 19.62 N - 2.887 Kg*v₂²/m
⇒ v₂ = 3.1517 m/s
We have TBC ≤ 35 N for v ≤ v₂ , that is, for v ≤ 3.1517 m/s
Combining the results obtained, we conclude that the range of allowable value is 2.816 m/s ≤ v ≤ 3.1517 m/s