Answer:
4592.57 lb
Explanation:
The missing diagram for this question is attached in the image below.
Given that:
the weight of the car = 2890 lb
At point A, the speed of the car [tex](V_A)[/tex] = 58 mi/hr
At point C, the speed of the car [tex](V_C)[/tex] = 18 mi/hr
To ft/s:
[tex](V_A)[/tex] = 58 mi/hr × 5280 ft/1 mi × 1 hr/3600 s
[tex](V_A)[/tex] = 85.07 ft/s
[tex](V_C)[/tex] = 18 mi/hr × 5280 ft/1 mi × 1 hr/3600 s
[tex](V_C)[/tex] = 26.4 ft/s
Between A to C, the total distance is;
[tex]S_{AC} = S_{AB}} + S_{BC} \\ \\ S_{AC} = 331 + \dfrac{\pi r}{2} \\ \\ S_{AC}= 331 + \dfrac{\pi \times 207}{2} \\ \\ S_{AC} = 656.154 \ ft[/tex]
Now, we need to determine the deceleration of the car using the formula:
[tex]V_C^2 = V_A^2 + 2 aS_{AC}[/tex]
[tex]26.4^2 = 85.07^2 + 2 a (654.154)[/tex]
[tex]696.96 = 7236.9049+ 2 a (654.154)[/tex]
[tex]696.96-7236.9049 = 2 a (654.154)[/tex]
[tex]-6539.9449 = 2 a (654.154)[/tex]
[tex]a= \dfrac{-6539.9449} {2(654.154)}[/tex]
a = -4.99 ft/s²
The velocity of the car as it passes via B
[tex]v_B^2 = v_A^2 + 2aS_{AB}[/tex]
[tex]v_B^2 = 85.07^2 + 2(-4.99 \times 331)[/tex]
[tex]v_B =\sqrt{ 85.07^2 + 2(-4.99 \times 331)}[/tex]
[tex]v_B =\sqrt{ 85.07^2 +3303.38}[/tex]
[tex]v_B =\sqrt{ 10540.2849}[/tex]
[tex]v_B =102.67 \ ft/s[/tex]
Along B, the car's acceleration is:
[tex]a_B = \sqrt{a^2 + (\dfrac{v_B^2}{r})^2}[/tex]
[tex]a_B = \sqrt{(-4.99)^2 + \dfrac{102.67^2}{207}^2 }[/tex]
[tex]a_B = 51.17 \ ft/s^2[/tex]
Finally, the total horizontal force F exerted = m[tex]a_B[/tex]
[tex]= (\dfrac{2890}{32.2}) \times 51.17[/tex]
= 4592.57 lb
