Problem 3/3 Page 129, Engineering Mechanics - Dynamics, Meriam and Kraige, 4th Edition
Given: A box of weight W has an initial speed of V0 up an incline making
an angle of q with the horizontal. The kinetic coefficient of friction
between the box and the incline is mk.
Find: The time required for the box to come to rest. The distance
traveled before the box comes to rest. The distance required for
the box
to slow to speed V1.
Parameter Values:
W = 80 lb
V0 = 30 ft/s
V1 = 15 ft/s
q = 20 degrees
mk = 0.25
1. Mechanical System: Box as it travels up the incline.
2. Free Body Diagram: Not shown. Would
include the box. The weight of the box acting vertically downward.
A normal force
acting perpendicular to the incline plane and a frictional force down
the incline plane. The frictional force is given by the kinetic coefficient
of
friction multiplied by the normal force. Coordinate axes adjacent
to the free body diagram show the t direction up the incline in the direction
of the motion and the n direction normal to the incline.
3. Equations:
S F = ( -W sin q
- mk N ) t + ( N - W cos
q ) n
m a = W/g ( at t + 0 n ) = W/g dv/dt t =
W/g dv/ds ds/dt t = W/g v dv/ds t
4. Solve:
Noting that there is no acceleration in the n direction:
N - W cos q = 0
N = W cos q
To investigate the question regarding time, we use the expression for
acceleration involving the derivative with respect to time:
-W sin q - mk
N = W/g dv/dt
Using the result for N:
-W sin q - mk
W cos q = W/g dv/dt
Dividing the equation by W and solving for dv/dt:
dv/dt = -g ( sin q + mk
cos q )
Separating variables:
dv = -g ( sin q + mk
cos q ) dt
Integrating both sides:
v = -g ( sin q + mk
cos q ) t + C
Recalling that v is V0 when t is zero, permits the evaluation of C:
v = -g ( sin q + mk
cos q ) t + V0
When the particle comes to rest at time tR, v is zero:
0 = -g ( sin q + mk
cos q ) tR + V0
Solving for tR:
tR = V0/g 1/( sin q + mk
cos q ) = 1.615 s
To investigate the questions regarding distance, we use the expression
for acceleration involving the derivative with respect to distance
traveled:
-W sin q - mk
N = W/g v dv/ds
Using the result for N:
-W sin q - mk
W cos q = W/g v dv/ds
Dividing the equation by W and solving for v dv/ds:
v dv/ds = -g ( sin q + mk
cos q )
Separating variables:
v dv = -g ( sin q + mk
cos q ) ds
Integrating both sides:
1/2 v2 = -g ( sin q + mk
cos q ) s + C
Recalling that v is V0 when s is zero, permits the evaluation of C:
1/2 v2 = -g ( sin q + mk
cos q ) s + 1/2 V02
When the particle comes to rest after traveling distance sR, v is zero:
0 = -g ( sin q + mk
cos q ) sR + 1/2 V02
Solving for sR:
sR = V02/2g 1/( sin q + mk
cos q ) = 24.2 ft
Determining the distance s1 to the point where the speed is reduced
to v1:
1/2 V12 = -g ( sin q +
mk cos q ) s1 + 1/2
V02
Solving for s1:
s1 = 1/2(V02 -V12)/g 1/( sin q
+ mk cos q ) = 18.17
ft