Recent content by sweet877

Thanks for your help!

F2 <------[ M ]---T---[ m ]----> F1 F2 = 20 N M = 2.0 kg m = 1.0 kg F1 = 30 N Find T, tension. So F1-F2 = (M+m)a. a=10/3 = 3.3 m/s^2 Fnet= Ma T-F2 = Ma T = Ma + F2 = 2.0(3.3) + 20 = 27 N However, I don't think this is right because I got a different answer when I used the other...

I drew a diagram, but somehow mistakenly thought that it is the scale exerting the force, and the elevator played no part in it. I get it now! Thanks very much for your help, lightgrav!

Oh wow...thanks!

A student stands on a bathroom scale in an elevator at rest on the 64th floor of a building. The scale reads 836 N. a) As the elevator moves up, the scale reading increases to 935 N, then decreases back to 836 N. Find the acceleration of the elevator. b) As the elevator approaches the 74th...

250 m and 400 m. But I just got that from common sense. Is there a "calculus way" to do it?

To contain oil spills, rectangular booms that have a cross-link to provide stability are used. The cross-link joins the long sides and is parallel to the short sides. What is the minimum total length of boom required to enclose an oil spill covering 100 000 m^2 of water if it can only be...

Oh OK...I get it now. Thanks!

Wait...wouldn't gravity be negative though?

I see...thanks for your help!

There is gravity...hmm... So the force the spider is exerting w/ 0 acceleration is mg = 2.1 X 10^-4 (9.8) = 2.06 X 10^-3 Fnet = ma = (2.1 X 10^-4)(a)

A 2.1 X 10^-4 kg spider is suspended from a thin strand of spider web. The greatest tension the strand can withstand without breaking is 2.0 X 10^-3 N. What is the maximum acceleration with which the spider can safely climb up the strand? Fnet = ma = (2.1 X 10^-4)(a) < 2.0 X 10^-3 a < 9.52...

Thanks for your help!

So, since F=\mu N, \mu = F/N = 9800 N/24500 N = 0.40?

Is the frictional force the opposite of the force that was applied to the car in order for it to stop?