# Ejercicios de friccion

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Friction

4-15. A horizontal force of 40 N will just start an empty 600-N sled moving across packed snow. After motion is begun, only 10 N is needed to keep motion at constant speed. Find thecoefficients of static and kinetic friction.
s = 0.0667; k = 0.016
4-16. Suppose 200-N of supplies are added the sled in Problem 4-13. What new force is needed to drag the sled atconstant speed?
N= 200 N + 600 N = 800 N; Fk = kN = (0.0167)(800 N); Fk = 13.3 N
4-17. Assume surfaces where s = 0.7 and k = 0.4. What horizontal force is needed to just start a50-N block moving along a wooden floor. What force will move it at constant speed?
Fs = s¬N = (0.7)(50 N) = 35 N ; Fk = s¬N = (0.4)(50 N) = 20 N
4-18. A dockworker finds that ahorizontal force of 60 lb is needed to drag a 150-lb crate across the deck at constant speed. What is the coefficient of kinetic friction?
; k = 0.400
4-19. The dockworkerin Problem 4-16 finds that a smaller crate of similar material can be dragged at constant speed with a horizontal force of only 40 lb. What is the weight of this crate?Fk = s¬N = (0.4)W = 40 lb; W = (40 lb/0.4) = 100 lb; W = 100 lb.
4-20. A steel block weighing240 N rests on level steel beam. What horizontal force will move the block at constant speed if the coefficient of kinetic friction is 0.12?
Fk = s¬N = (0.12)(240 N) ; Fk = 28.8 N.4-21. A 60-N toolbox is dragged horizontally at constant speed by a rope making an angle of 350 with the floor. The tension in the rope is 40 N. Determine the magnitude of the friction force andthe normal force.
Fx = T cos 350 – Fk = 0; Fk = (40 N) cos 350 = 32.8 N
Fy = N + Ty – W = 0; N = W – Ty = 60 N – T sin 350
N = 60 N – (40 N) sin 350;...