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The machine below in Figure 1.1 can be used to push boxes in a factory. The
dimensions shown are in centimeter (cm). The driver link 2 is at 125° and it is rotating
at 100 rpm in clockwise direction. Do the following.
Figure 1.1: A box-pushing system
(a) Use the Grashof’s Criterion to determine the relevant type of the four-bar
mechanism labeled ABCD. Can the driver link rotate fully?
(5 marks)
(b) Analyze the mobility of the machine. Where is a good place to locate the
actuator?
(3 marks)
(c) Use the position analysis to determine the angles 𝜽
𝟒
, 𝜽
𝟔
and 𝒓
𝟕
. The units are
in degrees () and m.
(12 marks)
(d) Use the velocity analysis to find the angular speed 𝜔
4
and linear velocity 𝑣
𝐸
.
Use rpm as the unit for angular speed. Positive rotation if it is in counter-
clockwise (CCW). How do you verify your answers?
(5 marks)
3
2
4
6
2
𝟏
𝟐
𝟑
𝟒
𝟓
𝟔
𝟒
𝟑
𝟔
5
𝟐
A
B
C
D
E
𝟐
2
𝟕
SOLUTION
POSITION:
VELOCITY:
clc
clear
close all
theta2real = 125; % deg
𝜽
𝟔
hold on
pusher = [C E]
plot(pusher(1,:), pusher(2,:), 'k-','linewidth',3)
% floatpoint = [B X C];
% plot(floatpoint(1,:), floatpoint(2,:), 'k-','linewidth',3)
axis equal
%% Velocity
disp('Myszka''s method')
w2 = - 100; % rpm CW
w2rps = w2*pi/
theta3 = theta3real; theta4 = theta4real; theta2 = theta2real;
w3 = - w2((L2sind(theta4 - theta2))/(L3*sind(gamma)))
w4 = - w2((L2sind(theta3 - theta2))/(L4*sind(gamma)))
w6 = - w4(L4cosd(theta4))/(L6*cosd(theta6))
v7 = - L4w4sind(theta4) + L6w6sind(theta6)
%% Check velocity
disp('Vector Method')
Mat34 = [L3sind(theta3) - L4sind(theta4);-L3cosd(theta3) L4cosd(theta4)];
V2 = L2w2rps[-sind(theta2);cosd(theta2)];
w34 = Mat34\V2;
w34 = inv(Mat34)*V2;
w3 = w34(1)
w4 = w34(2)
% Slider-crank
theta6 = - theta6;
Mat67 = [L6sind(theta6) 1;-L6cosd(theta6) 0];
V4 = L4w4[-sind(theta4);cosd(theta4)];
wv = inv(Mat67)*V4;
w6 = wv(1)
v7 = wv(2)
The Output:
BD = 118.22 cm
gamma = 71.522
theta3real = 41.664
theta4real = 113.19
theta6 = 17.196
L7 = 115.73 cm
B =
C =
D =
E =
points =
2
cos 𝜃
2
sin 𝜃
2
3
cos 𝜃
3
sin 𝜃
3
1 𝑥
4 𝑦
4
cos 𝜃
4
sin 𝜃
4
We also know that between BD, we can have another vector
ℎ
1
2
1 𝑥
4 𝑦
2
cos 𝜃
2
sin 𝜃
2
ℎ
2
2
ℎ
= tan
− 1
From here, use the geometry and trigonometry.
− 1
[
ℎ
2
3
2
4
2
ℎ
3
]
= cos
− 1
[
2
2
2
]
3
ℎ
ℎ
3
4
Next, we find the angle
4
− 1
[
ℎ
2
4
2
3
2
ℎ
4
]
= cos
− 1
[
2
2
2
]
4
ℎ
Then, with 𝜔
2
100
30
rps, get the angular velocities of 3 and 4.
2
3
1
4
2
2
−sin 𝜃
2
cos 𝜃
2
3
3
−sin 𝜃
3
cos 𝜃
3
4
4
−sin 𝜃
4
cos 𝜃
4
3
3
−sin 𝜃
3
cos 𝜃
3
4
4
−sin 𝜃
4
cos 𝜃
4
2
2
−sin 𝜃
2
cos 𝜃
2
[
3
sin 𝜃
3
4
sin 𝜃
4
3
cos 𝜃
3
4
cos 𝜃
4
]
3
4
2
2
2
2
[
] {
3
4
3
4
3
4
The second loop CDEF:
4
6
5
7
4
cos 𝜃
4
sin 𝜃
4
6
cos 𝜃
6
sin 𝜃
6
5
7
cos 113. 1863
sin 113. 1863
cos 𝜃
6
sin 𝜃
6
7
Use the y-component:
6
6
