Answer:
A) 13.80 ft/s^2
B) 1.714 Ib
Explanation:
Magnitude of acceleration center G
mass = W / g = 8 / 32.2 = 0.2484 Ib.s^2/ft
calculate the acceleration along x direction
A = ra
r = radius
a = angular acceleration
A = 6 in [tex]\frac{1 ft}{12 in}[/tex] * a
a= 2A
equation of the plane along the x-direction
w sin∅ - F = ma
8* sin40 - F = 0.2484 * a
hence F = 5.1423 - 0.2484 a
next find the moment of inertia along the z axis
I = 1/2 mr^2
= 1/2 * 0.2484 * (6/12)^2 = 0.03105 Ib.ft.s^2
Applying moment balance equation
F * r = inertia * a
(5.1423 - 0.2484 a)*0.5 = 0.03105 * 2A
2.57115 = 0.1863 A hence
A = 13.80 ft/s^2 ( acceleration of the cylinder )
B) Calculate the friction force exerted by the ramp on the cylinder
F = 5.1423 - 0.2484 A
= 5.1423 - 0.2484 ( 13.80 )
= 1.714 Ib
The magnitudes of the acceleration and the friction force are;
Acceleration = 13.8 ft/s²
Friction Force = 1.714 lb
The image of the solid homogeneous cylinder is missing and so i have attached it.
From the image we see that;
We are given;
We know that formula for weight is; W = mg
Thus; m = W/g
where g is acceleration due to gravity = 32.2 ft/s²
m = 8/32.2
mass; m = 0.2484 lb.s²/ft
Now, to get the acceleration along the x-axis, we will use the formula;
a = rα
where α is angular acceleration. Thus;
a = 0.5α
α = 2a ----- (eq 1)
Now, resolving forces along the x-direction gives;
W*sinθ - F = ma
Plugging in the relevant values;
8*sin 40 - F = 0.2484a
F = 8*sin 40 - 0.2484a -----(eq 2)
Now, moment of inertia of the cylinder along the z-axis is gotten from;
I = ¹/₂mr²
I = ¹/₂ × 0.2484 × 0.5²
I = 0.03105 lb.ft/s²
Taking equilibrium of moments we have;
F*r = I*α
Thus;
(8*sin 40 - 0.2484a)0.5 = 0.03105α
⇒ 2.57115 - 0.1242a = 0.03105α
⇒ 0.03105α + 0.1242a = 2.57115
From eq 1, α = 2a. Thus;
0.03105(2a) + 0.1242a = 2.57115
0.1863a = 2.57115
a = 2.57115/0.1863
a = 13.8 ft/s²
F = 8*sin 40 - 0.2484a
F = 5.1423 - 0.2484(13.8)
F = 1.714 lb
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