Math Group
Question:
kela000
Student
Community / Jr. College
Find the coefficient of kinetic friction between the flight bag and the floor.
A flight attendant pulls her 74.2 N flight bag a distance of 324m along a level airport floor at a constant velocity. The force she exerts is 37.9 N at an angle of 55.2 degrees above the horizontal.
Answers:
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krishna-agrawala Teacher
Graduate SchooleNotes Editor
Posted by krishna-agrawala on Thursday November 5, 2009 at 6:48 AMGiven:
Force exerted vertically by flight bag due to gravity = f1 = 74.2 N
Pulling force exerted = f2 = 37.9 N
Angle of pulling bag = A = 55.2 degrees.
Please not that distance moved has no impact on coefficient of kinetic friction.
The horizontal component of pulling action (f3) is acting on the bag to overcome the kinetic friction, and the vertical component (f4) is exerting an upward force on bag opposite to the gravitational pull (f1).
Thus net normal force acting on bag f5 = f1 - f4
f3 = f2*Cos A = 37.9*Cos 55.2 = 37.9*0.5707 = 21.6295 N
f4 = f2*Sin A = 37.9*Sin 55.2 = 37.9*0.8211 = 31.1197 N
f5 = f1 -f4 = 74.2 - 31.1197 = 48.0803 N
Coefficient of kinetic friction = f3/f5 = 21.6295/48.0803 = 0.4498
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neela Teacher
Graduate SchooleNotes Editor
Posted by neela on Thursday November 5, 2009 at 7:51 AMThe forces acting on the flight bag are:
Along the floor the frictional force and the component of the pulling force of 37.9N , which is 37.9N cos55.2 and in opposite direction.
The frictional force = m times total normal forces =-m( 72N-37.9N sin55.2)= -m*(72- 37.9sin55.2)N
The net force is zero as the flight bag is in constant velocity.
-m(72-37.7sin55.2)+37.9cos55.2=0
Therefore, m= 37.9cos55.2/(72-37.9sin55.2)
= 0.5291 is the coefficient of kinetic friction
