Datasheet

Page 32· Applied Robotics with the SumoBot
Figure 1-12
Individual Free Body
Diagrams
Since we already know that T = T
1
, we now know that T = W
1
. Next, calculate the sum
of the forces of the block on the table in both the x and y axis directions.
WN
0WN
0Fy
=
=−
=
∑
1S
S1
S1
S
X
Wf
fW
0fW
0fT
0F
=
=
=−
=−
=
∑
So, our key for solving this puzzle is that the normal force N is equal to the object's
weight, and the force of static friction f
S
is equal to the weight of the hanging object. We
already know the masses of both objects, and we also know that f
S
is the weight of W
1
,
and N is the weight of W. So, all that's left is to calculate the forces. To solve for this in
terms of force, first convert the weights to SI units of kg. Then, convert the masses to
newton forces, and finally, plug these values into the equation for the coefficient of
kinetic friction.
First, convert to SI units of kg:
kg1.2
g1000
kg1
g2100
kg7.0
g1000
kg1
g700
=×
=×