Engineering Dynamics Solutions

This document provides solutions for engineering dynamics problems, including equations of motion, cartesian coordinates, velocity, and pressure force. It covers topics such as cylindrical coordinates, position vectors, and constant forces. The solutions are explained step-by-step with relevant equations and calculations. The document also includes references for further study.

ENGINEERING DYNAMICS 1
Engineering Dynamics Solutions
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Engineering Dynamics Solutions.
4. Letting the cylindrical co-ordinates be:
Er is the unit vector in xy plane in radial direction.
Eꬾ is the unit vector in xy plane perpendicular to Er in the direction of increasing ꬾ.
K is the unit vector in z. Thus,
R = (x2 + y2)1/2 ; ꬾ = tan-1(y/x).
The position is: rop = x(t)i + y(t)j +z(t)k
= r(ꬾ)Er + z(t)k
Er = cosꬾi+sinꬾj= ∂ r
∂ r / ∂ r
∂ r
Eꬾ=
∂ r
∂ ꬾ /∂ r
∂ ꬾ
but ∂ r
∂ ꬾ = r(-sinꬾi+cosꬾj)
Velocity thus = vp = drop/dt=rer+rder/dt+zk+zdk/dt
When z-direction(k) fixed = dk/dt = 0
In figure 4.4, calculating the pressure force and coordinate x.
= mgsinꬾ
= 0.09×0.9×sin60
=0.070151
Also,
=0.01 × 0.1 ×sin 60
= 0.000866
X= x0 + v0t +1/2gsinꬾt2
0.07015= 0.2 + ½ sin60t2 + 4πt
0.07015 – 0.2 = 0.433t2 + 12.57t

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- 0.12985 = 0.433t2 + 12.57t
Also,
0.000866 = 0.5 + 1/2sin60t2 + πt
0.000866 – 0.5 = 0.433t2 + 3.14t
-0.49914 = 0.433t2 + 3.14t
- 0.12985 = 0.433t2 + 12.57t
0.36929 = 9.43t
T=0.0392
Solution to figure 4.15
Er is the unit vector in xy plane in radial direction.
Eꬾ is the unit vector in xy plane perpendicular to Er in the direction of increasing ꬾ.
K is the unit vector in z. Thus,
R = (x2 + y2)1/2 ; ꬾ = tan-1(y/x).
The position is: rop = x(t)i + y(t)j +z(t)k
= r(ꬾ)Er + z(t)k
Er = cosꬾi+sinꬾj= ∂ r
∂ r / ∂ r
∂ r
Eꬾ=
∂ r
∂ ꬾ /∂ r
∂ ꬾ
but ∂ r
∂ ꬾ = r(-sinꬾi+cosꬾj)
Velocity thus = vp = drop/dt=rer+rder/dt+zk+zdk/dt
When z-direction(k) fixed = dk/dt = 0
In figure 4.15, calculating the pressure force and coordinate x.
= mgsinꬾ
= 0.01×0.1×sin60

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=0.000866
X= x0 + v0t +1/2gsinꬾt2
Also,
0.000866 = 0.0005 + πt
0.000866 – 0.0005 = 3.14t
0.000366 = 3.14t
T= 0.0001166
5. Determine the velocity at positions B and C and get the pressure force on the
particle.
V = 4Q/πD2
But Q = ∫
0
D 2
u 2 πdr
Q = 2π ∫
0
D 2
urdr
= 2π × 0.25
=1.571
(4 × 1.571)/ π × 0.1×0.1
Velocity at B= 200.02m/s
Velocity at C
V = 4Q/πD2

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But Q = ∫
0
D 2
u 2 πdr
Q = 2π ∫
0
D 2
urdr
2π × 0.005
Q = 0.03142
Velocity = (4 × 0.03142)/ π×0.1×0.1
Velocity at C = 4.00m/s
P = 4T0/D
Pressure = 4 × 0.5/ 0.1
=20N/m2
5.15 Determine the velocity at positions B and C and get the pressure force on the
particle.
V = 4Q/πD2
But Q = ∫
0
D 2
u 2 πdr
Q = 2π ∫
0
D 2
urdr
= 2π × 0.25
=1.571
(4 × 1.571)/ π × 0.1×0.1
Velocity at B= 200.02m/s
Velocity at C
V = 4Q/πD2

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But Q = ∫
0
D 2
u 2 πdr
Q = 2π ∫
0
D 2
urdr
2π × 0.01
Q = 0.06283
Velocity = (4 × 0.06283)/ π×0.02×0.02
Velocity at C =199.9 m/s
P = 4T0/D
Pressure = 4 × 0.01/ 0.02
=2N/m2
6. Choose the cartesian coordinates from the figure provided hence,
Mx= -mgsin - μmgcosꬾ ꬾ
Giving x = - g (sin +μcos ) = - 5g/8ꬾ ꬾ
The time of upward motion of the brick is therefore,
T = x0/x = 5(5g/8) = 0.25s
The displacement of the particle becomes;
X = x0t +1/2xt2 = 5/8m
For t¿ t 1 x< 0, the equation of motion becomes
Mx = -mgsin + μmgcosꬾ ꬾ
X= -g(sin -μcos ) = -3g/8ꬾ ꬾ

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The displacement during the time interval are 0.25s and 0.5s
Thus, change in x = -1/2 ×3g/8×1/16
=-3/8 m
Thus, the displacement of the brick at t = 0.5s becomes
S = x + ∆ x = 5/8 – 3/8
= 0.25m
6.15 Choose the cartesian coordinates from the figure provided hence
Mx= mgsin - μmgcosꬾ ꬾ
Giving x = g (sin +μcos ) = 6g/8ꬾ ꬾ
The time of downward motion of the brick is therefore,
T = x0/x = 5(6g/8) = 0.5s
The displacement of the particle becomes;
X = x0t +1/2xt2 = 6/8m
For t¿ t 1 x< 0, the equation of motion becomes
Mx = mgsin + μmgcosꬾ ꬾ
X= g(sin -μcos ) = 2g/8ꬾ ꬾ
The displacement during the time interval are 0.5s and 1.0s
Thus, change in x = -1/2 ×2g/8×1/16
=-1/128 m
Thus, the displacement of the brick at t = 0.5s becomes
S = x + ∆ x = 6/8 – 2/8
= 0.5m

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7. Determine the constant force P and the equation of motion.
R = 0.6 sinꬾ
= 0.6 sin 30
=0.3cm
R* = 0.6 cosꬾꬾ*
= 0.6 cos 30 (3)
=1.559cm/s
R**= 0.6(cosꬾꬾ* - sin (ꬾ ꬾ*)2)
= 0.6 [co s 30(0) – sin 30 (32)]
= - 4.5cm/s2
Using the above time derivatives,
Αr = R* - Rꬾ2
= - 4.5 – 0.3(9)
=- 7.2cm/s2
A0 = Rꬾ* +2Rꬾ*
=0.3(0) + 2(1.559) (3)
= 9.354cm/s2
From the geometry shown in the figure, it is noted that the angle is halved to 150 for both
α and β. The force developed is therefore given by F = ks
= 100(0.6sin30 – 0.6sin 15)
=14.47N

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Equations of Motion By referring to the figure
+↑∑ fr = ma
N cos 15 – 14.47 – 0.2(9.81)
= 0.2(- 7.2)
N = 1.44N
+↓∑ f 0 = ma
FOA – 0.2(9.81) sin 15 – 23.87 sin 15
= 0.2(9.354)
=1.8708N
7.15Determine the constant force P and the equation of motion.
R = 0.6 sinꬾ
= 0.6 sin 20
=0.2052cm
R* = 0.6 cosꬾꬾ*
= 0.6 cos 20 (3)
=1.6914cm/s
R**= 0.6(cosꬾꬾ* - sin (ꬾ ꬾ*)2)
= 0.6 [co s 20(0) – sin 20 (32)]
= - 1.847cm/s2
Using the above time derivatives,
Αr = R* - Rꬾ2
= - 1.847 – 0.2052(9)
=- 3.6938cm/s2

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A0 = Rꬾ* +2Rꬾ*
=0.2052(0) + 2(1.6914) (3)
= 10.1484cm/s2
From the geometry shown in the figure, it is noted that the angle is halved to 100 for both
α and β. The force developed is therefore given by F = ks
= 100(0.6sin20 – 0.6sin 10)
=10.10N
Equations of Motion By referring to the figure
+↑∑ fr = ma
N cos 10 – 10.10 – 0.2(9.81)
= 0.2(- 5.2)
N = 1.04N
+↓∑ f 0 = ma
FOA – 0.2(9.81) sin 10 – 23.87 sin 10
= 0.2(8.263)
=1.6526N

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References
Massachusetts Institute of Technology. (n.d). Rigid Bodies: Rotational and Transitional
Motion Rolling without Slipping. Available from,
http://web.mit.edu/8.01t/www/materials/Presentation_W11D1.pdf
Wilson. K., Addie. G., Clift. R., & Sellgren. A. (2006). Review of Fluid and Particle
Mechanics. Springer. Available from, http://www.springer.com/978-0-387-23262-1

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