Recent content by ammcl30

That makes so much more sense thank you! So theoretically if after I calculate the static friction and it was less than the applied force, I could solve the kinetic friction using Newtons 2nd law and kinematics. Correct?

If there was no person pushing the box up the incline then the box would slide down because there are no forces to act on it going up the ramp. The forces that push the box down would be gravity and static friction.

Thank you for answering but this leads me to a question. Why do we only account for the static friction and not the kinetic friction? Don't both of these factor into the overall movement of an object?

A 10kg block is at rest on the floor, and then a force of 30 N pushes on the block horizontally. If the coefficient of static friction between the block and the floor is μs= 0.33 and the coefficient of kinetic friction between the block and the floor is μk= 0.20, what is the acceleration of the...

Okay so F= .15 * 283.98= 42.59 N of maximum static friction. Obviously I could just pick the correct answer at this point, but I'm sure there is a way to calculate the minimum force. My first reaction would be to calculate the F= μM because that would account for just the objects weight against...

Okay so my normal force would be : (30)(9.8)cos (15°) = 283.982 Then from this I would plug my normal force into F=μN?

Box pushed up an incline So the component would be (30)cos15= 28.97. Cosine because I defined the incline as the x direction.

1. A 30kg box is placed on a 15° slope and a person pushes on the box up the slope to keep it from sliding. If the coefficient of static friction is 0.15, what is the minimum force that the person must exert to keep the box in place? Answer Choices: A. 273N B. 241N C. 33.5 N D. 64.7 N E. 294 N...