solved: a schematic of a clutch-testing machine is shown. | studysoup
a schematic of a clutch-testing machine is shown. the steel shaft rotates at a constant speed v.
an axial load is applied to the shaft and is cycled from zero to p. the torque t induced by
the clutch face onto the shaft is given by
t 5 f p (d 1 d)
4
 where d and d are defined in the figure and f is the coefficient of friction of the clutch face. the
shaft is machined with sy 5 120 kpsi and sut 5 145 kpsi. the theoretical stress-concentration
factors for the fillet are 3.0 and 1.8 for the axial and torsional loading, respectively.
 assume the load variation p is synchronous with shaft rotation. with sy 5 0.3, find the
maximum allowable load p such that the shaft will survive a minimum of 106
 cycles with a
factor of safety of 3. use the modified goodman criterion. determine the corresponding factor
of safety guarding against yielding.
engr 3341 probability theory and statistics prof. gelb week 5 homework solutions warm-up problems from textbook: section 3.3 problem 9: in this problem, we would like to show that the geometric random variable is memoryless. let x ▯ geometric(p). show that p(x > m+ljx > m) = p(x > l); for m;l 2 f1;2;3;:::g we can interpret this in the following way: remember that a geometric random variable can be obtained by tossing a coin repeatedly until observing the ﬁrst heads. if we toss the coin several times and do not observe a heads, from now on it is like we start all over again. in other words, the failed coin tosses do not impact the distribution of waiting time from now on. the reason for this i
the clutch might be smelly, but is it slipping or doing anything else untoward. clutches are a regularly reported problem on one-tonne utes, which suggests most are borderline in capacity. if there is a problem with the functioning of the clutch, approach mazda again about replacing it. if there isn't a problem think about replacing the clutch with a heavy-duty one that's more capable of handling your towing needs. ![How do you test a coil?](https://static.3dstockphoto.com/thumb/00/00/61/90/uwx619072.jpg)  
Photo Credit: Courtesy of 3DStockPhoto ([Do You image](https://3dstockphoto.com/Do-You-uwx619072))The only safe way to test for spark is to use a spark plug tester tool. If a coil problem is suspected, measure the coil's primary and secondary resistance with an ohmmeter. If either is out of specifications, the coil needs to be replaced. A coil can be easily bench tested with a digital 10 megaohm impedance ohmmeter. the value of head loss due to friction using chezy formula with constant c= 60 for a fluid, flowing through a 30 cm diameter and 70 m long pipe. velocity of fluid inside the pipe is 5 m/s. time to have a mechanic do some hands-on testing! https://carkiller.com/scottykilmer/qa/mechanics/ Here's the diagram you requested. All the best!  

Greg  

![I need a serpentine belt routing diagram for a - c528ac5.gif](http://i.fixya.net/uploads/images/c528ac5.gif) [image] ryan\_blything:

i have difficulty understanding why (since ultimately, rts should be dependent on keyboard press rather than frame rate).

builder experiments are structured around a drawing loop and event cycle that is driven by the screen refresh rate. so the keyboard is checked once … 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. there are so many things that can cause rough idle, can you give more information? did the car recently overheat? is there any fluids leaking on the engine? are you getting any smoke or steam from the exhaust? when checking for steam, remember to wait until the engine has run for awhile as condensation can build up in your exhaust during the night. if you cannot tell if it is smoke or steam, hold a blank sheet of paper behind the tail pipe, if it is oil smoke, then oil dots will show up on the paper.

some things you can do and check;
check the spark plug wires, make sure they are tightly connected and not chaffed, broken or burnt.
check your pcv valve for excessive oil or clogging, (it will be in a rubber housing or hose leading from your valve cover to your throttle body/carburetor)
i don't know the engine setup in that model, but if you have a distributor cap, check it for cracks and wear, also check the rotary cap.
check all rubber vacuum hoses on the engine for cracking.
if you can afford it, change the spark plugs, plug wires, rotor and rotary cap, these parts all need to be changed over time and your car is getting older now. 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.