Answer:
a = 3.608 ft/s²
T = 5.977 kip
Explanation:
Angle of inclined track is not given. It is 5 degree as shown in attached image 1.
Given data:
Weight of boxcar A = Wα = 20,000 lb
Weight of boxcar B = Wb = 30,000 lb
Coefficient of kinetic friction = υk = 0.5
angle of inclined track = 5 degree
Value of g = 32.2 ft/s²
Solution:
These boxcars are moving therefore,
∑Fₓ = ma
∑[tex]Fy = 0[/tex]
Car A
Use ∑[tex]Fy = 0[/tex]
Forces acting in Y-axis are weight and its counter N (as shown in Free body diagram). Weight is Negative (acting downward) while N is positive (acting upward)
∑[tex]Fy =[/tex] Nα - Wα*Cos(∅) = 0
∑[tex]Fy =[/tex] Nα - 20,000*Cos(5°) = 0
∑[tex]Fy =[/tex] Nα - 20,000*0.9962 = 0
∑[tex]Fy =[/tex] Nα - 19,924 = 0
Nα = 19,924
Use ∑Fₓ = maₓ
Forces acting in x-axis are Fa, T and horizontal component of weight and force having acceleration a (as shown in Free body diagram). Horizontal component of weight and Tension (T) are Negative (acting in negative X-axis direction) while Fa is positive (acting in positive X-axis direction)
∑Fₓ = maₓ
(0.5*19924) - 20,000*Sin(5°) - T = (20,000/32.2)*a
9962 - 1744 - T = 621.12*a
8218 - T = 621.12*a .................. Eq(1)
Both cars
∑Fₓ = maₓ
Forces acting are; horizontal component of weights, Fa and force having acceleration a (as shown in Free body diagram).
[tex](0.5*19924) - 20,000*Sin(5°) - 30,000*Sin(5°) = (20,000/32.2)/a + (30,000/32.2)/a[/tex]
9962 - 1744 - 2616 = 621.12/a + 931.68/a
5602 = 1552.8 / a
⇒ a = 3.608 ft/s²
put a = 3.608 ft/s² in equation (1)
⇒ T = 8218 - 621.12*3.608
T = 8218 - 2240.79
T = 5977.21 lb
T = 5.977 kip
