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Need help solving this question with D'Alembert principle!

[From: ] [author: ] [Date: 11-05-08] [Hit: ]
and thus no acceleration normal to the road.Hence:N = m*g*cos(theta)Forces parallel to the road (define forward as +x):Component of weight: -m*g*sin(theta)Drag: -DTraction: +FNewtons second law for these forces:F - m*g*sin(theta) - D = m*aSolve for F:F = m*(a + g*sin(theta)) + DNOW...we substitute previous expressions for a and theta:a = (vf^2 - vi^2)/(2*d)theta = arctan(S)Thus:F = m*((vf^2 - vi^2)/(2*d) + g*sin(arctan(S))) + DWe can simplify the trig(arctrig) combination as:sin(arctan(S)) = S/sqrt(1 + S^2)Thus:F = m*((vf^2 - vi^2)/(2*d) + g*S/sqrt(1 + S^2) ) + DSince we also desire work done, we want to evaluate:W = integral (F dot dx)OR.......




CALL THE SLOPE ANGLE theta.

We can get the slope angle from the slope S (given as value 1/5) using arctangent.

theta = arctan(S)


We will desire to access all trigonometry of theta...so that is why we desire to express as an angle instead of a slope.


Forces perpendicular to the road (define up as +y):

Component of gravity: -m*g*cos(theta)
Normal force: N

These forces add up to zero, because the road is assumed straight, and thus no acceleration normal to the road.

Hence:
N = m*g*cos(theta)


Forces parallel to the road (define forward as +x):
Component of weight: -m*g*sin(theta)
Drag: -D
Traction: +F

Newton's second law for these forces:
F - m*g*sin(theta) - D = m*a



Solve for F:
F = m*(a + g*sin(theta)) + D



NOW...we substitute previous expressions for a and theta:
a = (vf^2 - vi^2)/(2*d)
theta = arctan(S)


Thus:
F = m*((vf^2 - vi^2)/(2*d) + g*sin(arctan(S))) + D


We can simplify the trig(arctrig) combination as:
sin(arctan(S)) = S/sqrt(1 + S^2)


Thus:
F = m*((vf^2 - vi^2)/(2*d) + g*S/sqrt(1 + S^2) ) + D



Since we also desire work done, we want to evaluate:
W = integral (F dot dx)

OR...since force is constant (special case):
W = F*d

Thus:
W = d*(m*((vf^2 - vi^2)/(2*d) + g*S/sqrt(1 + S^2) ) + D)



To get POWER, we also need the TIME INTERVAL.
t = 2*d*(vf - vi)/(vf^2 - vi^2)

By definition of AVERAGE power:
P_avg = W/t

Thus:
P_avg = d*(m*((vf^2 - vi^2)/(2*d) + g*S/sqrt(1 + S^2) ) + D)/(2*d*(vf - vi)/(vf^2 - vi^2))

Simplify:
P_avg = (m*((vf^2 - vi^2)/(2*d) + g*S/sqrt(1 + S^2) ) + D)/(2*(vf - vi)/(vf^2 - vi^2))
P_avg = (m*((vf^2 - vi^2)/(2*d) + g*S/sqrt(1 + S^2) ) + D)*(vf^2 - vi^2)/(2*(vf - vi))


Summary:
A: F = m*((vf^2 - vi^2)/(2*d) + g*S/sqrt(1 + S^2) ) + D

B: W = d*(m*((vf^2 - vi^2)/(2*d) + g*S/sqrt(1 + S^2) ) + D)

C: P_avg = (m*((vf^2 - vi^2)/(2*d) + g*S/sqrt(1 + S^2) ) + D)*(vf^2 - vi^2)/(2*(vf - vi))



DATA:
m = 1350 kg
g = 9.8 Newtons/kilogram; PLEASE...remember where this came from. It is not an arbitrary number...it has physical meaning as Earth's gravitational field.
d = 100 m
S = 0.2
vi = 2.5 m/s
vf = 12 m/s
D = 750 Newtons


Results:
A: F = 4275 Newtons
B: W = 42750 Joules
C: P_avg = 26074 Watts

D: you didn't ask for it, but it is important when sizing engines. Maximum power: P_max = F*v_max ... thus P_max = F*vf

P_max = vf*(m*((vf^2 - vi^2)/(2*d) + g*S/sqrt(1 + S^2) ) + D)
P_max = 51294 Watts
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