Warm Up (10/26/15) Friction

Transcription

1 Warm Up (10/26/15) Friction trained sea lion slides from rest with constant acceleration down a 3.0 m long ramp into a pool of water. If the ramp is inclined at an angle of 23 above the horizontal and the coefficient of kinetic is 0.26, how long does it take for the sea lion to make a splash in the pool. Fnet = ma; system: sea lion Fg + Fk,fr + = ma

2 Warm Up (10/26/15) Friction trained sea lion slides from rest with constant d = vo t acceleration + ½a t 2 down a 3.0 m long ramp into a pool of water. If the ramp is inclined Fg + Fk,fr at an + angle = ma of 23 above the horizontal and the coefficient of kinetic is 0.26, how long does it take for the sea lion to make a splash in the pool. Fnet = ma; system: sea lion ( mgcos + ),N + (mgin Fk,fr), E = ma ( mgcos + ) = 0 & (mgin Fk,fr) = ma = mgcos (mgin μk ) = m(2d t 2 ) 0

3 Warm Up (10/26/15) Friction trained sea lion slides from rest with constant d = vo t acceleration + ½a t 2 down a 3.0 m long ramp into a pool of water. If the ramp is inclined at an angle of 23 above the horizontal and the coefficient of kinetic is 0.26, how long does it take for the sea lion to make a splash in the pool. ( mgcos + ),N + (mgin Fk,fr), E = ma ( mgcos + ) = 0 & (mgin Fk,fr) = ma Fnet = ma; system: sea lion = mgcos (mgin μk ) = m(2d t 2 ) Fg + Fk,fr + = ma 0

4 Daily Quiz #4 (10/27/15) Friction block is placed on a rough surface inclined relative to the horizontal as shown in the figure. The incline angle is increased until the block starts to move. If the coefficient of static μ s is 0.25, what is the weight of the block? Fnet = ma; system: block Fg + Fs,fr + = ma Fs,fr Fg

5 Daily Quiz #4 (10/27/15) Friction block is placed on a rough surface inclined relative to the horizontal as shown in the figure. The incline angle is increased until the block starts to move. If the coefficient of static μ s is 0.25, what is the weight of the block? Fnet = ma; system: block Fg + Fs,fr + = ma ( mgcos + ),N + (mgin Fs,fr), E = ma 0 ( mgcos + ) = 0 & (mgin Fs,fr) = 0 mgin = Fs,fr Fs,fr mgin = μs mg = μs in Fg

6 4.9 tatic and Kinetic Friction Calculate the force acting on the crate when F app = 0, F app > 0 but crate does not move, and when F app > 0 and crate slides. F app 70.0 kg 70.0 kg μ static = 0.40 F app Ffr (N) Fstatic = μstatic max F static Fsliding = μsliding Fstatic μ sliding = 0.30 Fstatic max F app (N) Fsliding F app F app 70.0 kg

7 Lab 7 Determining the Coefficient of Friction Fkfr = μk

8 4.9 tatic and Kinetic Friction Friction has its basis in surfaces that are not completely smooth. Fnet = + = 0.

9 4.9 tatic and Kinetic Friction Friction has its basis in surfaces that are not completely smooth. F app Fstatic + = 0. ND Fstatic + Fapp = 0.

10 4.9 tatic and Kinetic Friction Friction has its basis in surfaces that are not completely smooth. F app Fstatic + = 0. ND Fstatic + Fapp = 0.

11 4.9 tatic and Kinetic Friction Friction has its basis in surfaces that are not completely smooth. + = 0. Fapp Fsliding Fsliding < Fapp

12 Warm Up (10/28/15) Friction person has a choice of either pushing or pulling a sled at a constant velocity, as the drawing illustrates. Friction is present. If the angle is the same in both cases, does it require less force to push or to pull the sled? Fkinetic = μ To pull, because the upward component of the pulling force reduces the normal force and, therefore, also reduces the force of kinetic acting on the sled.

13 Case 2 Warm Up (10/30/15) Friction box has a weight of 150 N and is being pulled across a horizontal floor by a force that has a magnitude of 110 N. The pulling force can point horizontally, or it can point above the horizontal at an angle. When the pulling force points horizontally, the kinetic al force acting on the box is twice as large as when the pulling force points at the angle. Find Case 1 mg = 2mg 2FT in 2FT in = mg Fk,fr = μk = μk (mg) in = mg 2FT Fk,fr = μk = in 1 (mg 2FT) = μk (mg FT in) μk (mg) =2 [ μk (mg FT in)] F N mg F T in