Answer:
Usually, a solution can have several criteria and constraints. Even though all are important, some criteria are more important than others. The same holds true for constraints. But what do you do if it's impossible for a solution to cover every criterion while avoiding every constraint? In cases like this, you can use prioritization. Listing criteria and constraints based on priority shows the relative importance of each. You will need to prioritize the criteria and constraints for each sub-problem so that you can design a solution for each one individually. Prioritization can help you compare two different possible solutions. For example, the criterion that cars travel at 15 mph through the neighborhood might be a higher priority than the constraint that homeowners are only willing to spend $10,000 on this issue. If this is the case, you would want to generate solutions that also follow the priority in mind. All criteria are important, but engineers must sometimes make a trade-off, which is a compromise or change in one or more criteria or constraints so that they can be met at the same time. This is where prioritization comes in handy as it helps determine the trade-offs. A solution that is doing a better job of meeting one criterion may result in not completely meeting another criterion. Prioritization will help you choose which solution to go with.
Explanation:
I got this from quizlet :)
Answer:
it would take him 1 minute to run 304.5 meters and 1 second to run 5.075 meters
Explanation:
Inertia is the correct answer!
Since the ladder is standing, we know that the coefficient
of friction is at least something. This [gotta be at least this] friction
coefficient can be calculated. As the man begins to climb the ladder, the
friction can even be less than the free-standing friction coefficient. However,
as the man climbs the ladder, more and more friction is required. Since he
eventually slips, we know that friction is less than what's required at the top
of the ladder.
The only "answer" to this problem is putting lower
and upper bounds on the coefficient. For the lower one, find how much friction
the ladder needs to stand by itself. For the most that friction could be, find
what friction is when the man reaches the top of the ladder.
Ff = uN1
Fx = 0 = Ff + N2
Fy = 0 = N1 – 400 – 864
N1 = 1264 N
Torque balance
T = 0 = N2(12)sin(60) – 400(6)cos(60) – 864(7.8)cos(60)
N2 = 439 N
Ff = 439= u N1
U = 440 / 1264 = 0.3481