the column is built up by gluing the two boards together. | studysoup
the column is built up by gluing the two boards together. determine the maximum normal stress on the cross section when the eccentric force of p = 50 kn is applied.
1. introduction: rough surface in contact i. surface of engineering materials: generals  surfaces have chemical properties, which are different from bulk virgin material. o most metallic surfaces in air have oxide layers, which change the chemical and mechanical properties of the surfaces.  surfaces have mechanical properties, which are different from bulk material. o machining or sliding contact makes strain hardened (or work hardened) layer and highly deformed layer, whose strength are different from bulk.  real engineering surfaces are rough. mirror like smooth surface afm image
that depends on a lot of things. in particular, you would need to find the electric susceptibility of the paper. the field produced by static charge is inverse-square. the polarization of the paper varies directly with the external field, with a proportionality constant equal to the electric susceptibility. the dipole field created by the polarized paper is inverse-cube. overall, the force between the ruler and the paper will vary with the inverse fifth power of distance. to get any more detailed, you would need to know the geometry of the situation, and the charge distribution. given: (i) xy = 4.5cm

(ii) yz = 5 cm

(iii) zx = 6 cm

to construct: a xyz

steps for construction

step 1 : first we will draw a line segment xy of length 4.5cm



step 2 : we have to draw an arc , taking x as center and radius 6cm



step 3 : we have to another arc, with radius 5cm and center at y and mark the intersection of arcs as z.



step 4 : we have to join xz and xy



this is the required triangle coulomb's law:f=kq1q2r2, where k is the coulomb's constant and r is the separation between the two charges. the magnitude of the net force:fnet=fx2+fy2k = 9 × 109 n·m2/c2the force due to charge q1 is:f1a=kq1qar1a2=(9×109)(21.4×10-6)(13.5×10-6)(80.1×10-2)2[readmore]f1a = 4.05 nthe force due to charge, q2:f2a=kq2qar2a2=(9×109)(30.6×10-6)(13.5×10-6)[(80.1×10-2)2+(80.1×10-2)2]2f2a = 2.9 nforce due to charge, q3:f3a=kq3qar3a2=(9×109)(57.3×10-6)(13.5×10-6)(80.1×10-2)2f3a = 10.9 nthe net force along the x-axis:fx=f1a-f2acos(45°)=4.05-(2.9)cos(45°)fx = 2.0 nthe net force in the y-direction:fy=f3a-f2acos(45°)=10.9-(2.9)cos(45°)fy = 8.85 nthe magnitude of the net force:fnet=fx2+fy2=2.02+8.852fnet = 9.07 nthe magnitude of the net force on the charge at point a is 9.07 n. check out the intersect function.
here is one approach:
% generate a rectangular boundry
rectx = [0.1 0.1 0.9 0.9];
recty = [0.1 0.5 0.5 0.1];
rect = polyshape(rectx,recty);

% initialize a figure for display purposes
figure(1);clf
titles = [...
 "all lines";...
 "partially within";...
 "exclusively within"];
for i = 1:3
 subplot(1,3,i)
 hold on
 title(titles(i))
 plot(rect)
 xlim([0 1])
 ylim([0 1])
end

% define the number of line segments to generate
nlines = 25;

% initialize a structure to store the lines (if you need the data later)
lines = struct();

% generate each line, calculate it's intersection conditions, and plot
for i=1:nlines
 
 % generate points for a random line segment
 lines(i).points = rand(2);
 
 % section the line into parts at intersection with the rectangle
 [in, out] = intersect(rect, lines(i).points);
 
 % plot all the lines that were generated
 subplot(1,3,1)
 line(lines(i).points(:,1),lines(i).points(:,2))
 
 % plot lines that are partially within the boundary
 if isempty(in)==0
 subplot(1,3,2)
 line(lines(i).points(:,1),lines(i).points(:,2))
 end
 
 % plot lines that are exclusively within the boundary
 if isempty(out)==1
 subplot(1,3,3)
 line(lines(i).points(:,1),lines(i).points(:,2))
 end
end
here is the output: coefficient of friction is defined as the ratio of the force of friction acting between two objects and the force which is pushing them together. the value of the coefficient of friction varies from zero to one. there are two types of the coefficient of friction: static coefficient of friction kinetic (also known as sliding) coefficient of friction. therefore, the coefficient of friction depends on the nature of the surface. you don't say that the area must be a rectangle. the true largest area would be a circle, with the circumference equal to 400m and the area would be 12,732.4 m^2. however, evidently, the area is restricted to a rectangle, in which case the maximum area is a square, and the side of the square will be 
400/4 = 100 m. the area is 100x100 = 10,000 m^2.
algebraically,
let x = one pair of opposite sides
(400-2x)/2 = 200-x = the other pair of opposite sides

area = a(x) = x(200 - x) = -x^2 + 200x
x-coordinate of vertex = -b/2a = 200/2 = 100
a(100) = 10,000

that's it! ;) the most direct test is to measure pressure in the spinal fluid. this is done with a spinal tap and pressure measuring column. the fluid column is measured, and then the fluid is drained off. just draining the fluid can relieve symptoms. analysis of the fluid would rule out more serious conditions. this procedure is easily performed in a doctor's office or emergency room. see more... update in case anyone encounters this admittedly specialized use case and needs a workaround.

the following code avoids the removechild error seen in the javascript console.

fundamentally, what is being done here is creating a placeholder for the gauge via pn.column() and then appending the gauge … The given case is SAS (two sides and included angle) so use the Law of Cosines:
  

\(\displaystyle{b}^{{2}}={a}^{{2}}+{c}^{{2}}-{2}{a}{c}{\cos{{B}}}\)
  

\(\displaystyle{b}^{{2}}={5}^{{2}}+{3}^{{2}}-{2}{\left({5}\right)}{\left({3}\right)}{\cos{{92}}}°\)
  

\(\displaystyle{b}^{{2}}={34}-{30}{\cos{{92}}}°\)
  

\(\displaystyle{b}=\sqrt{{{34}-{30}{\cos{{92}}}°}}\)
  

b ~ 5.92 the graph of this equation is a line.  we know that because each term is either a constant (1.5) or the product of a constant and the first power of a variable (d, 0.8t).
since we have two variables (d and t), the line sits in a two-dimensional plane, with a vertical axis and a horizontal axis.  d stands alone in the equation, so its value depends on the value oft.  therefore, the vertical--or dependent--axis will represent values of d, and the horizontal--or independent--axis will represent values of t.
at this point, then, we can set up a coordinate plane with a vertical axis labeled d and a horizontal axis labeled t.  all of the points on our plane have a coordinate (t, d).
to graph a line, we need two points.  since the question asks for the axial intercepts anyway, we can use those.
we'll start with the d-intercept, where the line crosses the d-axis.  all of the points on the d-axis have a t-coordinate of zero, so to find where the line crosses the axis, we substitute zero for t in our equation and solve.
d = 1.5 + 0.8 × t
d = 1.5 + 0.8 × (0)
d = 1.5 + 0
d = 1.5
so our d-intercept has the coordinates (0, 1.5), and we have our first point and our first intercept.
similarly, to find the t-intercept, we note that all of the points on the t-axis have a d-coordinate of zero, so we set d equal to zero and solve.
d = 1.5 + 0.8 × t
0 = 1.5 + 0.8 × t
0 - 1.5 = 1.5 - 1.5 + 0.8 × t
-1.5 = 0.8 × t
-1.5 ÷ 0.8 = (0.8 ÷ 0.8) × t
-1.875 = 1 × t
-1.875 = t
and the d-intercept is (-1.875, 0).
now we have both axial intercepts, so that part of the question is complete.  we can plot those two points and draw the line that passing through them (extending on, at either side, to infinity).
all that remains is to find the turning points.  a turning point, also called a local maximum or local minimum, is where the value of a function goes from decreasing to increasing or from increasing to decreasing.  (the graph shown at http://en.wikipedia.org/wiki/file:parabola\_with\_focus\_and\_directrix.svg has a minimum at the origin, for example.)
a line has a constant slope.  to say that differently, the angle between any two points on a line is the same as the angle between any two other points on the line.  if a line is increasing in one location (like our line), it is increasing everywhere, and it is increasing everywhere at the same rate.  therefore, a line has no turning points.