Dynamics III Friction and Inclines

FBD

Example A box with a mass of 60.0kg is being pulled by two forces as show below. Find the net acceleration of the box. Which way is the box moving? 12.0 N θ = 30.0°

What will the scale read if the elevator is at rest? Snowman mass = 50 kg, Scale Reading is Normal Force N mg ???????? 490 (N)

What will the scale read if the elevator has an upward acceleration of 2 m/s2 ? Snowman mass = 50 kg (he doesn’t melt) N mg ???????? 590 (N) Scale reads 590(N) increasing speed on the way up OR decreasing speed on the way down

What will the scale read if the elevator is moving at a constant velocity? Snowman mass = 50 kg N mg ???????? 490 (N) Scale reads 490(N) for constant velocity on the way up or way down (and when at rest)

What will the scale read if the elevator has a downward acceleration of 3 m/s2 ? Snowman mass = 50 kg N mg ???????? 340 (N) Scale reads 340(N) increasing speed on the way down OR decreasing speed on the way up

What will the scale read if the elevator has a downward acceleration of 9.8 m/s2 ? Snowman mass = 50 kg N mg ???????? 0 (N) Scale reads 0(N) increasing speed on the way down OR decreasing speed on the way up

Friction Force Friction force is caused by surfaces sliding over each other Symbol: Ff Direction: Opposite to the motion of the object Magnitude: If object is sliding, 𝐹 𝑓 = 𝜇 𝑘 𝐹 𝑛 where 𝜇 𝑘 = coefficient of kinetic friction Magnitude: if object is not sliding 𝐹 𝑓 = 𝜇 𝑠 𝐹 𝑛 where 𝜇 𝑠 = coefficient of static friction

Example A box of books is being dragged across the ground ( = 0.81) by a rope (FA = 170 N, 33 from horizontal). If m = 15 kg, calculate ax. Fx = FAcos  Ff = max ax = (FAcos  Ff) We would be done except we don’t know Ff How do we get Ff? m = 15 kg ax = ? FA = 170 N   = 33  = 0.81 FA Fg FN Ff m

FBD on an incline