Recent content by tanh^-1(x)

Ummm... Now how do I do problems #2 and #3? I made two more equations but they only account for the smaller box of weight 32kg, Net Force X = 0 = -32*9.8*sin(27.76) + Fn*.756 (5) Net Force Y = 0 = -32*9.8*cos(27.76) + Fn (6) So through (6) I get Fn= 277.5 N, plugging that into (5) doesn't equal...

It worked! Thanks for the 301 N! I was calculating Fn * .756 not by .435, so I got the 1st answer right!

HI, Yes, thank you for the fn*.756! However, I still get -159.3 as my answer, which is also wrong. How do you calculate friction force to be 301 N? I calculated it to be 693 N, which is not what you said. Maybe you mean: Net Force X = -80*9.8*sin(angle) + Fpull + Fn*.435 = 0 (3) This way I get...

Homework Statement You are lowering two boxes, one on top of the other, down the ramp shown in the figure by pulling on a rope parallel to the surface of the ramp. Both boxes move together at a constant speed of 19.0cm/s . The coefficient of kinetic friction between the ramp and the lower box...