Transcript Home Work #1 - Hong Kong University of Science and Technology

Home Work #1
Due Date: 11 Feb, 2010
(Turn in your assignment at the mail box
of S581 outside the ME general office)
The solutions must be written on single-side A4 papers only.
HW 1-Problem #1
The beam is subjected to the parabolic loading. Determine an equivalent
force and couple system at point A.
1
1
8 31 8
F   wdx   8 x dx  x  (kN )
3 0 3
0
0
F=2.667kN
2
MRA=0.667kN.m
1
1
8 31 8 41 2
M RA  M o   w(1  x)dx   (8 x  8 x )dx  x  x  (kN )
3 0 4 0 3
0
0
2
3
HW 1-Problem #2
Two couples act on the frame. If the resultant
couple moment is to be zero, determine the
distance d between the 500-N couple forces.
D=1.663m
M1=500*0.9*sin60
M2=-500*(0.9+d)*sin60
M3=-750*(1.8+d)*3/5
M4=750*[(1.8+d)*3/5+1.2*4/5]
M1+M2+M3+M4=0
D=1.663m
HW 1-Problem #3
4F
F
2F
2F
4F
2F
F
2F
The five ropes in figure can each take 1500N without breaking. How
heavy can W be without breaking any?
7F=W
4F=1500N
W=7*1500/4=2625N
HW 1-Problem #4
The man in figure weighs 800N. He pulls down
on the rope, raising the 250-N weight. He finds
that the higher it goes, the more he must pull to
raise it further. Explain this, and calculate and
plot the rope tension T as a function of θ. What
is the value of the tension, and the angle θ,
when the man can lift it no further? Neglect the
sizes and weights of the pulleys.
2T * cos  250
Tmax  800
 max  81
HW 1-Problem #5
Two small balls A and B have masses m and 2m, respectively.
They rest on a smooth circular cylinder with a horizontal
axis and with radius R. They are connected by a thread of
length 2R. Find the angles θ1 and θ2 between the radii and
the vertical line OC for equilibrium, as well as the tension in
the thread and forces exerted by A and B on the cylinder.
Assume that the balls are very small and that the tension is
2 R 180
constant.
1   2 
Fa
F F
θ1
mg
θ2
2mg
R


 114 .59
Fa*sin θ1=F*cos θ1
Fa*cos θ1+F*sin θ1=m
Fb Fb*sin θ =F*cos θ
2
2
Fb*cos θ2+F*sin θ2=2m
sin θ1= 2sin θ2
θ1=84.73
θ2=29.86
F=mgsin θ1=0.9958mg N
Fa=mg cos θ1 =0.0919mg N
Fb=2mg cos θ2 =1.7345mg N