Problem 6/75 Page 458, Engineering Mechanics - Dynamics, Meriam and Kraige, 4th Edition
Given: Uniform pole of mass M and length L balanced in a vertical position on a horizontal surface. The coefficient of friction between the pole and the surface is mk. A force P is applied to a point on the rod a distance L/3 above the ground.
Find: The initial acceleration of the upper end of the rod just an instant after the application of the force.
Parameter Values:
M = 50 kg = Mass of pole.
L = 6 m = Length of pole.
mk = 0.3 = Kinetic coefficient
of friction.
P = 500 N
1. Mechanical System: Pole at the instant of application of the force.
2. Free Body Diagram: Shows the pole at the instant of application (note that the pole is still vertical). Shows a weight force acting vertically downward through the center of the rod. Shows a normal force acting vertically upward at the bottom of the rod. Shows the force P acting to the right at a point L/3 up from the bottom of the rod. Shows a friction force acting to the left at the bottom of the rod. The friction force is given by mk times the normal force. The x axis is directed horiztonally to the right and the y axis is vertically upward.
3. Equations:
a) S F = ( P - mk
N ) i + ( N - M g ) j
S MG = -L/6 j X
P i + -L/2 j X ( -mk
N i + N j )
= ( P L/6 - L/2 mk
N ) k
b) m aG = M ( aC + w
X ( w X rG/C ) + a
X rG/C)
C is the contact point with the ground. This point is known to
move horizontally so any acceleration it has is in the x direction.
The pole is released from rest so the initial angular velocity is 0.
The pole rotates in the xy plane so the angular acceleration is known to
be in the z direction. The vector from the contact point to the mass center
is of length L/2 and is directed vertically upward.
m aG = M ( aC i + a
k X L/2 j )
= M ( aC - a L/2
) i
IG a = 1/12 M L2 a
k
(Using the mass moment of inertia for a slender rod).
To determine the acceleration of the upper end of the rod we are going
to need another two point formula.
aA = aC + w
X ( w X rA/C ) + a
X rA/C
A is the upper end of the rod, a distance L vertically above the contact
point C. Noting again that the angular velocity is zero at the instant
of interest, we obtain:
aA = ( aC - a
L ) i
4. Solve:
P - mk N = M ( aC
- a L/2 )
N - M g = 0
P L/6 - L/2 mk N = 1/12 M L2
a
aA = ( aC - a
L ) i
From the second equation:
N = M g
From the third equation:
a = 2 P/(M L) - 6 mk
g/L = 0.390 rad/s2
From the first equation:
aC = a L/2 + P/M
- mk g = 8.23 m/s2
From the fourth equation:
aA = ( aC - a
L ) i
aA = ( 5.89 m/s2) i