A dustbin of mass 10 kg standing on the horizontal ground is pushed by a
horizontal force of 60 N. If the coefficient of friction is 0.4, will the dustbin move? (If the Applied is more than the Friction...). If so, what is the acceleration of the dustbin?
Answers
Answer:
First consider the horizontal or x-axis:
F
net
x
=
T
2
x
−
T
1
x
=
0.
Thus, as you might expect,
T
1
x
=
T
2
x
.
This gives us the following relationship:
T
1
cos
30
°
=
T
2
cos
45
°
.
Thus,
T
2
=
1.225
T
1
.
Note that
T
1
and
T
2
are not equal in this case because the angles on either side are not equal. It is reasonable that
T
2
ends up being greater than
T
1
because it is exerted more vertically than
T
1
.
Now consider the force components along the vertical or y-axis:
F
net
y
=
T
1
y
+
T
2
y
−
w
=
0.
This implies
T
1
y
+
T
2
y
=
w
.
Substituting the expressions for the vertical components gives
T
1
sin
30
°
+
T
2
sin
45
°
=
w
.
There are two unknowns in this equation, but substituting the expression for
T
2
in terms of
T
1
reduces this to one equation with one unknown:
T
1
(
0.500
)
+
(
1.225
T
1
)
(
0.707
)
=
w
=
m
g
,
which yields
1.366
T
1
=
(
15.0
kg
)
(
9.80
m/s
2
)
.
Solving this last equation gives the magnitude of
T
1
to be
T
1
=
108
N
.
Finally, we find the magnitude of
T
2
by using the relationship between them,
T
2
=
1.225
T
1
, found above. Thus we obtain
T
2
=
132
N
.