Recent content by katye

and just for anyone who tries to use this in the future, the answer is the normal force you found in the first question*cos(angle)*distance :) best of luck!

nevermind i got it! thanks!

sorry, I'm trying to solve for how much work the rope is doing now and i thought the equation would just be frictional force=applied force so (frictional coeff)*(Normal force)*(distance) but i was wrong, wrong, wrong. so .8(130.229)(19.2) because 130.229 was what i found for my first answer for...

oh my gosh. thank you so much. i know it seems really simple, but that problem has been killing me. i really appreciate it!

alright that make since. but my homework service keeps telling me i';m wrong. do you see any mistakes F(x)=0=Xcos(20.2) - (.8*17.2*9.8) and i get X=143.68567 then i use that for F(y)=0=N+(17.2)(-9.8)+ (143.68567sin(20.2)) and i get N=118.9456

i've done that but i can't figure out how to solve for the applied force's x and y components because i can't set the sum of the forces in the x direction to zero because the box is moving right?

a cart loaded with bricks has a total mass of 17.2 kg and is pulled at constant speed by a rope. the rope is inclined at 20.2 degrees above the horizontal and the cart moves 19.2 m on a horizontal floor. the coeffiecient of kinetic friction between the ground and the cart is .8. The acceleration...