Recent content by xr250h

Alright so I changed my way of calculating Ff to Ff = u*ma = 19.6 N for part b. And for part c, doubling the acceleration, I get 39.12 N. Thank you so much for helping me! Time for sleep now, it's 1 AM here :eek: . Thanks again!

Ok I found out that the normal force is equal to the applied force, so I got Ff = \muma. Then, I used that to plug in for Ff. So I got \muma - mg = 0. I then solved for a and got 16.3 m/s^2. For part b, I know I could either use Ff=mg or Ff= u*ma and use my new found acceleration (both...

Ok So I've got Ff - Fg = m*ay and Ff = \mu*Fn. But I don't know where my normal vector would go? Does it go in the same direction as my acceleration or opposite of Fg?

Block and cart friction problem - SOLVED Homework Statement You and your best pal make a friendly bet that you can place a 2.0-kg box against the side of a cart, and that the box will not fall to the ground, even though you guarantee to use no hooks, ropes, fasteners, magnets, glues, or...

Homework Statement A bolt comes loose from underneath an elevator that is moving upward at a constant speed of 6.0 m/s. The bolt reaches the bottom of the elevator shaft in 3.0 s. a.) How high above the bottom of the shaft was the elevator when the bolt came loose? b.) What is the...