Download Final examination: Dynamics

Final examination: Dynamics
Hiroki Okubo
July 17 2013
z
A
R
θ
F
b
O
2b
B
y
ω
r
x
2F
Figure 2: Problem 3
force L which he exerts on the package during
the acceleration interval. (20)
Solution. We apply the equation of motion in
the vertical direction to get R − L − M g = M a
and L − mg = ma where a = g/4. Simultaneous solution gives R = (M + m)(a + g) =
5(M + m)g/4 N and L = 5mg/4 N.
Figure 1: Problem 1
1. Two forces act on the rectangular plate as
shown. Reduce this force system to an equivalent force-couple system acting at point O.
Then determine the resultant of the system,
expressed as a single force if possible, with its
line of action. (20)
Solution. The resultant force R is R =
(−F + 2F )k = F k. The moment about point
O is M O = bi × (−F )k + 2bj × (2F )k =
F b(4i + j). An approach in determining the
line of action of R is to use the vector expression r × R = M O where r = xi + yj + zk
is a position vector running from point O to
any point on the line of action of R. Substituting the vector M O and carrying out the
cross product result in (xi + yj + zk) × F k =
F b(4i + j). Thus, the desired line of action
is given by yi − xj = b(4i + j), y = 4b and
x = −b.
3. Small objects are released from rest at A and
slide down the smooth circular surface of radius R to a conveyor B. Determine the expression for the normal contact force N between the guide and each object in terms of θ
and specify the correct angular velocity ω of
the conveyor pulley of radius r to prevent any
sliding on the belt as the objects transfer to
the conveyor. (20)
Solution. By using the polar-coordinate
form, when the object on the circular surface
the equations of motion are mg cos θ = mRθ̈
and −N + mg sin θ = −mRθ̇2 . Integrating the
equation of motion in the tangent direction to
obtain as a function of θ yields
Z
2. A man with a mass of M kg holds a package
with a mass of m kg as he stands within an
elevator which briefly accelerations upward at
a rate of g/4. Determine the force R which the
elevator floor exerts on his feet and the lifting
θ̇
θ̇dθ̇
=
0
θ̇2
2
=
g
R
Z
θ
cos θdθ
0
g
sin θ
R
Substituting the value of θ̇ into the equation
1
z
B
2r
A
y
x
Figure 3: Problem 4
of motion in the normal direction gives N =
mg sin θ + mRθ̇2 = 3mg sin θ. The conveyor
pulley must turn √
at the rate Rθ̇ = rω for θ =
π/2, so that ω = 2gR/r.
4. A smooth homogeneous sphere of mass m and
radius r is suspended by a wire AB of length
2r from point B on the line of intersection of
the two smooth vertical walls at right angles
to one another. Determine the reaction R of
each wall against the sphere. (20)
Solution.
Solution. The vertical distance between
point
√
B and the center of the sphere is 7r. With
unit vectors i, j and k in the x-, y-, and zdirections, the zero summation of forces for
equilibrium yields the vector equations Rx i +
Ry j + T − mgk = 0 where Rx and Ry are
the reaction force in the x and y directions
and T = T is the tension of the wire. The
tension can be described as
√ T = T n where
n = −(1/3)i − (1/3)j + ( 7/3)k. Equating
the coefficients of the√i-, j- and k-terms
to
√
zero gives √T = 3mg/ 7, Rx = mg/ 7 and
Ry = mg/ 7.
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