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(mech103)[2006](f) sol~PPSpider^midterm_10513.pdf
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MECH 103 Midterm Examination
Fall Semester 2006

6:30 pm C 8:30 am 18 Oct 2006
Student Name:_________________
Student ID:___________________
Student Signature:______________
1 Kinematics: Relative Motion (20pts)
The crank AB is rotating with a constant counterclockwise angular velocity of 1000 rpm (revolutions per
minute). The length of the connecting rod BC = 100 mm.

(a)
Determine the acceleration of point B in terms of normal and tangential components,

(b)
Determine the acceleration of point B in terms of Cartesian components,

(c)
Determine the velocity vector of point B,

(d)
Determine the velocity vector of point C.




x

C
a
B r

x
A
The angular velocity of crank AB is
= 1000 rpm (2 rad/rev) (min/ 60sec) = 104.72 rad/s

The acceleration in the tangential direction is simply zero:
at = 0,
The acceleration in the normal direction can be calculated as follows:
an = (0.05 m) (104.72 rad/s)2 = 548.31 m/s2

(a)
aB= 548.31 en m/s2

(b)
aB= (548.31 m/s2)(cos 45 i C sin 45 j) = (387.7 i - 387.7 j) m/s2

(c)
vB = rAB = (0.05 m) (104.72 rad/s)= 5.236 m/s vB = 5.236 (-cos 45 i C sin 45 j) = (-3.702i -3.702 j ) m/s

(d)


= 45, From the law of sine, = sin.1(r sin / a) =20.7048 From the relative motion equation for the velocities of the point B and point A, v B = v A +k rB / A (*) From the relative motion equation for the velocities of the point B and point C, vB = vC +BC k rB / C (**)
The velocity of point C has only the component in the y direction:
vc = vc j

rB/A = 0.05 (-sin i + cos j ) = -0.0354 i + 0.0354 j
rB/C = 0.1 (-sin i -cos j ) = -0.0354 i - 0.0935 j

Let Eqn (*) = Eqn (**)
v + k r = v + k r
AB / ACBC B / C
0 +104.72k (.0.0354i + 0.0354j) = vc j +BC k (.0.0354i . 0.0935j)
. 3.7071i . 3.7071j = v j + 0.0935 i . 0.0354 j = 0.0935 i + (v . 0.0354 )j
C BC BC BCC BC
This vector equation can be used to find two unkowns: BC and vc BC = -39.6471, vc C0.0354 BC = -3.7071, vc = 0.0354(-39.6471)C3.7071= -5.11 m/s
vc = -5.11 j m/s
2 Newtons law (20pts)
At the instant shown in the following figure, the 11,000-kg airplanes velocity is v = 270 i (m/s). The forces
acting on the plane are its weight W, the thrust T = 110 kN, the lift L = 260 kN, and the drag D =34 kN. (The
x axis is parallel to the airplanes path.) Determine the airplanes acceleration vector (magnitude and angle).


Summation of all the 4 forces: From the above figure,
oo
x direction: Fx = T cos15 . D .W sin15
oo
y direction: Fy = L + sin15 .W cos15 According to Newtons 2nd law: F = ma
oo
Fx = T cos15 . D .W sin15 = max
oo
Fy = L + T sin15 .W cos15 = may a = (ax, ay) = (4.03, 16.75) m/s2

22 2
a
=
ax + ay =17.23 m/s
a = tan.1(16.75/ 4.03) = 76.47o
3 Orbital Mechanics (30pts)
Assume that the International Space Station (ISS) i