Question

a. Place the crate on top of the ramp with the angle at 20^o.  Draw a free body diagram showing all the forces and their components.

Free Body Diagram- Show Hide Friction Ice (no friction) Wood Vectors Force Vectors Sum of Forces Walls Brick Bouncy More Controls Object Position 8.9 meters -10.0 10.0 h = 3.4 m Ramp Angle 20.0 degrees 0.0 80.0 - 10 -8 -6 -4 -2 Ometers ->)) O on off Angle = 20.0 Ν. Applied Force Ο Reset All Small Crate 100 kg H=0.3 Hs=0.5 Clear O Record Playback

b. If the crate does not go down the ramp, which two forces are responsible for keeping the ramp at rest and preventing it from sliding down?

c. Slowly increase the angle until the block JUST starts to move.  What is the angle when the block JUST moves?

d. For the angle when the block JUST moves, calculate μ_s (Use the forces on your free body diagram to do this)

e. Compare the calculated value with the given value for μ_s for the crate.

Answer 1

o=20° , m= lookg, Uke 0-3, lls = 0.5 Griven, a) Free body diagram: N of a Em mg sinos mgwso 6 If the coate does not go down the samp static friction force is solely responsible in this case. ts=lls N and N= mg coso fs = els mg coso = 0.5x 100 kg x 9.8 m/s2 x cos(200) fs = 460.5N and masino = lookg x 9.8 m/s2 x sin (200) = 335N fs> mgsino Block will not slidey © 0=9. (when mation & Just startst until the block is at nest , static friction force works but the moment motion starts kinetic friction staots

to act: so in ar direction: mg sino = fs mg sino = Mis mg costo tano a las = 0.5 o = tan-1 (0.5) = 26. 6° Ans when we incr. angle up to 26.6°, fs stops acting and motion starts fx acts Surther. d) us=9 me fs = mg sino = 100x9.8 x sin (26 6) = 438.8 and N = mg coso = 100x9.88 Cos (26 6) = 876-3N fs = us N Ms = fs = 438.8 = 10.5li @ It will be same because we calculated N 876.3 one from other and vice versa.