Question

A machine in an ice factory is capable of exerting 3.00 x 10^2 N of force to pull a large block of ice up a slope. The block weighs 1.22 x 10^4 N. Assuming there is no friction, what is the maximum angle that the slope can make with the horizontal if the machine is able to complete the task?

Answer 1

In order for the block to be pulled up the ramp, the force of the pulling, FA, has to be greater than or equal to the force of the ice sliding down the ramp, Fgx or Wx.

looking at the force summation, Newton's second law, we can see
\[\sum F = F_A - F_{gx} =ma\]
and we know that the acceleration of the block has to be greater or equal to zero (the acceleration can equal zero if the block has some velocity prior to hittng the ramp)

\[ma≥0\]
therfore

\[F_A - F_{gx} ≥0\]
\[\quad F_{gx} = W \sin \theta\]
\[ F_A ≥ W \sin \theta\]
switching the sin theta over to the right means we flip around the inequality
\[\sin \theta ≤ \frac{W}{F_A}\]
\[ \theta≤\arcsin \Big(\frac{W}{F_A}\Big)\]

theta must be less than or equal to the the arcsine of the weight divided by the pulling force of the machine.

Answer 2

I prefer to use it this way, but thanks, I got the idea! (:\[\theta \le \sin^{-1} \left(\frac{W}{F_A}\right)\]

Answer 3

I'm mindful of that since it can also mean
\[\sin^{-1}\theta = \frac{1}{\sin \theta}=\csc \theta\]

(obviously it wouldn't in this case because of the leading steps, I know)
but I'm pretty sure your notation is what graphing calculators use for the symbol on the inverse sine button; Wolfram says it's common too, so mine seems to just be mostly old habits over use :P Sorry for the confusion.

Answer 4

I hate using cosecants! :p
But thanks, I've got the answer (;