Problem 2.51

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Problem 2.51 [Difficulty: 4]
Ff τ A⋅=
x, V, a
M g⋅
Given: Data on the block and incline
Find: Initial acceleration; formula for speed of block; plot; find speed after 0.1 s. Find oil viscosity if speed is 0.3 m/s after
0.1 s
Solution:
Given data M 5 kg⋅= A 0.1 m⋅( )2= d 0.2 mm⋅= θ 30 deg⋅=
From Fig. A.2 μ 0.4
N s⋅
m2
⋅=
Applying Newton's 2nd law to initial instant (no friction) M a⋅ M g⋅ sin θ( )⋅ Ff−= M g⋅ sin θ( )⋅=
so ainit g sin θ( )⋅= 9.81
m
s2
⋅ sin 30 deg⋅( )×= ainit 4.9
m
s2
=
Applying Newton's 2nd law at any instant M a⋅ M g⋅ sin θ( )⋅ Ff−= and Ff τ A⋅= μ
du
dy
⋅ A⋅= μ V
d
⋅ A⋅=
so M a⋅ M dV
dt
⋅= M g⋅ sin θ( )⋅ μ A⋅
d
V⋅−=
Separating variables dV
g sin θ( )⋅ μ A⋅
M d⋅ V⋅−
dt=
Integrating and using limits M d⋅
μ A⋅− ln 1
μ A⋅
M g⋅ d⋅ sin θ( )⋅ V⋅−
⎛⎜⎝
⎞
⎠⋅ t=
or V t( )
M g⋅ d⋅ sin θ( )⋅
μ A⋅ 1 e
μ− A⋅
M d⋅ t⋅−
⎛⎜⎝
⎞
⎠⋅=
At t = 0.1 s V 5 kg⋅ 9.81× m
s2
⋅ 0.0002× m⋅ sin 30 deg⋅( )⋅ m
2
0.4 N⋅ s⋅ 0.1 m⋅( )2⋅
× N s
2⋅
kg m⋅× 1 e
0.4 0.01⋅
5 0.0002⋅ 0.1⋅
⎛⎜⎝
⎞
⎠−−
⎡⎢⎣
⎤⎥⎦×=
V 0.1 s⋅( ) 0.404 m
s
⋅=
The plot looks like
0 0.2 0.4 0.6 0.8 1
0.5
1
1.5
t (s)
V
 (m
/s
)
To find the viscosity for which V(0.1 s) = 0.3 m/s, we must solve 
V t 0.1 s⋅=( ) M g⋅ d⋅ sin θ( )⋅
μ A⋅ 1 e
μ− A⋅
M d⋅ t 0.1 s⋅=( )⋅−
⎡⎢⎣
⎤⎥⎦⋅=
The viscosity µ is implicit in this equation, so solution must be found by manual iteration, or by any of a number of classic
root-finding numerical methods, or by using Excel's Goal Seek
Using Excel: μ 1.08
N s⋅
m2
⋅=