A horizontal conveyor belt system has a coefficient of kinetic friction of 0.65. The motor driving the system can deliver a maximum force of 2.5 X 10^6 N. What maximum total weight can be placed on the conveyor system?
Answers (1)
Get a ruler in your hands. Measure things until you start to understand how a ruler works. Measure some stuff and figure out where the center is. Say you measure a book and it's 7/8" thick. You look at your ruler and see that every eighth is divided into two sixteenths, so obviously half of 7/8" is going to be 7/16". If you write that out you have 1/2 x 7/8 = 7/16. And you notice that 1/2 is divided into 2/4 and then into 4/8 and so on, so you can convert anything to anything by multiplying all the numbers on top and then all the numbers on bottom.
Other rulers are divided into 10 and 100 parts. But an inch is still an inch, so anything on one ruler can be translated to the other ruler. A half inch on one ruler is 5/10 or 50/100 on the other. An eighth inch is just 12.5 marks when you have 100 marks per inch. A metric ruler divides an inch into 25.4 parts, so a half inch would be 12.7 of those parts. Pretty simple, isn't it? Practice this a bit and people will think you went to wizard school.
Coefficient of kinetic friction is defined as force to keep an object moving divided by weight.
u = f/w
0.65 = 2.5 X 10^6 N/w The rule is you can do any valid operation on both sides of an equation and it will still be equal. Multiply by w.
0.65w = 2.5 X 10^6 Divide by 0.65.
w = 3.846 x 10^6 N <-- ANSWER