Recent content by fajoler

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Alright so I did relate that Work is equal to the change in kinetic energy which is also equal to the magnitude of Force times the magnitude of the change in distance.By doing so, I get that the initial Kinetic Energy equals mg(0.04x + 0.1)xThe equation for u(x) = 0.04x + 0.1. The distance...

Oh wow... I'm sorry that was a stupid move on my part. For some reason I forgot that the normal force depends on gravity AND the direction of the gravitational force... Thanks so much I appreciate it!

Homework Statement A luggage handler pulls a suitcase of mass 17.1 kg up a ramp inclined at an angle 24.0 above the horizontal by a force of magnitude 146 N that acts parallel to the ramp. The coefficient of kinetic friction between the ramp and the incline is 0.257. The suitcase travels a...

Hey for the third problem you were having trouble with, I somehow ended up with g/2 as my answer. Could you tell me where exactly you messed up so I can see where I messed up as well? I'm not quite sure... Thanks!

ah youre right.so would the equation for y then be y = -(h/b)x + h?If I use this as my equation for y, and plug it into the integral for inertia, I come up with an answer of 1/6 Mb^2.this is the correct answer. thanks for walking me through this problem. I greatly appreciate it!

so y = mx + b and we can calculate that y = (-h/b) x assuming the point of rotation is aligned at the origin right?which means dA = (-h/b) x dxNow since r is the distance from the origin, or the point of rotation, can we say that r is equal to x and dx is equal to dr?then now that we know that...

Homework Statement A metal sign for a car dealership is a thin, uniform right triangle with base length b and height h. The sign has mass m. What is the moment of inertia of the sign for rotation about the side of length h? Homework Equations I = \intr^{2}dm The Attempt...

thanks so much. it turns out the problem i was having was the miscalculation of the radius. stupid mistakes like misreading directions can lead to a lot of trouble... thanks again!

thanks so much. it turns out the problem I was having was with the radius. A stupid misread direction can lead to a lot of trouble... Thanks again!

Homework Statement A mass of 13.0 , fastened to the end of an aluminum wire with an unstretched length of 0.50 is whirled in a vertical circle with a constant angular speed of 130 . The cross-sectional area of the wire is 1.3×10−2 . The Young's modulus for aluminum is Pa.Calculate the...