Specific ASTB Question (physics related)

[E6286]

A 300 pound weight rests on a hydraulic surface with an area or 376 square inches. What force is needed to act upon the smaller surface having an area of 36 sq. inches? Also, how far would you have to push the small surface to make the weight raise 1 inch?

I understand the first part. The Force on the small side would be 28.8lbs, however, how do I determine how far I would need to push the small side to make the larger side rise one inch?


Question 2:

If a swimmer is swimming at 3mph across a river to a dock and the river is flowing downstream at 6mph, how does one figure out:

a)the angle the swimmer needs to swim upstream to hit the dock?

b)the amount of time it will take to get to the dock. Actually, I think this part is easier. Correct me if I am wrong but if the swimmer is swimming into a 6mph current at 3mph than his total progress is at 3mph so you just break that into feet per second and multiply it by the width of the river.

 

[E6286]

Okay, I got the first problem. You use A1 x D1 = A2 X D2 to get

(376)(1) = (36)(D2)
= (376/36)
= 10.4 inches

Anything I am missing?

 

[asise]

Part 2 is poorly worded. If the swimmer is swimming slower than the current, than he won't ever reach the dock (poor bugger) - assuming he is starting from directly across the river. If instead the question means that his relative velocity is 3 mph while crossing the river, than that would mean he is in fact swimming at 6.7 mph at an angle upstream. Now he can reach the dock and the angle he must swim at is 63.4 degrees. Use SOH CAH TOA for that (sine is opposite over hypotenuse, etc...) There weren't any distances so it could take him all day if it's a particularly wide river. You would also want to take into account the ecosystem - are there pirahannas for example? Those little things could affect his swimming speed part way through.
Hope that helps.
This is a vector question - the key is to break down the velocity of the object into its components and use trig to figure out each component.

 

[asise]

And for part I ...
the force (in lbs) would be 28.7
Pressure is the same on both sides so:
300/376 = F/36
F = 28.7

 

[E6286]

asise: If he is swimming agains the current upstream at 3 mph and the river is moving downstream at 6 mph wouldn't he be moving at 3 mpr across? 3+(-6) = 3. Also, how did you calculate that angle. I know SOH CAH TOA and I figured that was how you use here but how did you determine which you needed to use and how did you get the angle?

 

[WarPoet]

asise: If he is swimming agains the current upstream at 3 mph and the river is moving downstream at 6 mph wouldn't he be moving at 3 mpr across? 3+(-6) = 3.

Keep in mind that adding a positive number to a negative number is the same as subtracting a negative number from a positive number. (3+(-6) = 3-6 = -3). This means that even if he were swimming directly up stream he would still be going downstream at 3mph. So, unless he starts out farther upstream (along the bank), or has a positive velocity relative to the downstream flow, he has no hope of making it to the dock no matter which angle he chooses to swim at. In fact, the larger the angle (relative to the flow of the river), the faster he will go downstream. (90 degrees -- heading straight toward the dock -- he will be going downstream at 6mph)

Re-read the question from the source and post it word for word...I suspect your interpretation of the question may be the hang-up. If you did post word for word...the error is in the problem and it's unsolvable.

 

[asise]

exactly,

The guys not swimming at 3 mph - so I assumed that 3 mph was the speed of his progress across the river while swimming at the correct angle. with 3 and 6 as two sides of a triangle I got 6.7 as his actual swimming speed. and Tan -1 (6/3) gives you the 64 degree angle.
As is the question is no good - had to tweak it to get an answer.

 

[kmac]

Boy, I'm glad I'm not taking the ASTB! That requires ....brain cells... and math....