Recent content by p0ink

An 18.8kg box is released on a 38.0o incline and accelerates down the incline at 0.281m/s2. What is the magnitude of the friction force impeding its motion. F = m*g*cos(38) sum forces parallel to the plane m*g*sin(38) - (mu)*(F) = m*a, or m*g*sin(38) - (mu)*m*g*cos(38) = m*a masses cancel...

ahhh... i see. the pressure was what was throwing me off. i kept thinking p=f for some odd reason. thanks!

And tank A would have an equal force on the base of the tank, since the side forces are not significant.

it would be less for tank B, because the sides are pushing with an upwards force. for tank C, it would more, since the sides of the tank are pushing with a downward force.

here is my thinking for the forces on the bottom of each tank, which made me come to those answers. A: F=m*g B: m*g - the upward force of the sides of the tank C: m*g + the downward force of the sides of the tank

i don't see why my answers: LGTFG and LGTTG can't be correct?

Tank A |...| |..h2o...| |..h2o...| |_h2o__| Tank B \.../ .\..h2o../ ..\.h2o./ ...\h2o/ Tank C .../...\ ../.h2o.\ ./..h2o..\ /__h2o__\ The three tanks shown above are filled with water to an equal depth. All the tanks have an equal height. Tank A has the middle...