Example 8.3 Balancing Act Goal Apply the conditions of equilibrium and illustrate the use of different axes for calculating the net torque on an object. Problem A woman of mass m = 51.4 kg sits on the left end of a seesaw -- a plank of length L = 4.43 m, pivoted in the middle as in Figure 8.6. (a) First compute the torques the seesaw about an axis that passes through the pivot point. Where should a man of mass M = 72.7 kg sit if the system (seesaw plus man and woman) is to be balanced? (a) L (b) Find the normal force exerted by the pivot point if the plank has a mass of mpl - 2.00 m 11.6 kg. mg Mg (c) Repeat part (b), but this time compute the torques about an axis through the left end of the plank. (b) Figure 8.6 (a) Two people on a see-saw. (b) Free body diagram for the plank. Strategy In part (a), apply the second condition of equilibrium, Et = 0, computing torques around the pivot point. The mass of the plank forming the seesaw is distributed evenly on either side of the pivot point, so the torque exerted by gravity on the plank, Taravity, can be computed as if all the plank's mass is concentrated at the pivot point. Then Tgravity is zero, as is the torque exerted by the pivot, because their moment arms are zero. In part (b), the first condition of equilibrium, EF = 0, must be applied. Part (c) is a repeat of part (a) showing that choice of a different axis yields the same answer. Exercise 8.3 Hints: Getting Started | I'm Stuck Suppose a 29.7 kg child sits 2.00 m to the left of center on the same seesaw. A second child sits at the end on the opposite side, and the system is balanced. (a) Find the mass of the second child. mchild 2 kg (b) Find the normal force acting at the pivot point. Fn
Angular Momentum
The momentum of an object is given by multiplying its mass and velocity. Momentum is a property of any object that moves with mass. The only difference between angular momentum and linear momentum is that angular momentum deals with moving or spinning objects. A moving particle's linear momentum can be thought of as a measure of its linear motion. The force is proportional to the rate of change of linear momentum. Angular momentum is always directly proportional to mass. In rotational motion, the concept of angular momentum is often used. Since it is a conserved quantity—the total angular momentum of a closed system remains constant—it is a significant quantity in physics. To understand the concept of angular momentum first we need to understand a rigid body and its movement, a position vector that is used to specify the position of particles in space. A rigid body possesses motion it may be linear or rotational. Rotational motion plays important role in angular momentum.
Moment of a Force
The idea of moments is an important concept in physics. It arises from the fact that distance often plays an important part in the interaction of, or in determining the impact of forces on bodies. Moments are often described by their order [first, second, or higher order] based on the power to which the distance has to be raised to understand the phenomenon. Of particular note are the second-order moment of mass (Moment of Inertia) and moments of force.
![Example 8.3 Balancing Act
Goal Apply the conditions of
equilibrium and illustrate the use of
different axes for calculating the net
torque on an object.
Problem A woman of mass m =
51.4 kg sits on the left end of a
seesaw -- a plank of length L = 4.43
m, pivoted in the middle as in Figure
8.6.
(a) First compute the torques
the seesaw about an axis that
passes through the pivot point.
Where should a man of mass M =
72.7 kg sit if the system (seesaw
plus man and woman) is to be
balanced?
(a)
L
(b) Find the normal force exerted
by the pivot point if the plank has a
mass of mpl
- 2.00 m
11.6 kg.
mg
Mg
(c) Repeat part (b), but this time
compute the torques about an axis
through the left end of the plank.
(b)
Figure 8.6 (a) Two people on a see-saw. (b) Free body diagram for the
plank.
Strategy In part (a), apply the second condition of equilibrium, Et = 0, computing torques around the pivot
point. The mass of the plank forming the seesaw is distributed evenly on either side of the pivot point, so the
torque exerted by gravity on the plank, Taravity, can be computed as if all the plank's mass is concentrated at
the pivot point. Then Tgravity is zero, as is the torque exerted by the pivot, because their moment arms are
zero. In part (b), the first condition of equilibrium, EF = 0, must be applied. Part (c) is a repeat of part (a)
showing that choice of a different axis yields the same answer.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3a74cc31-62f4-4982-89c3-ce8d7d2cfbb2%2Ff3d657f1-cc11-4d90-9c7a-b94031bd7d78%2Fg53s1q_processed.png&w=3840&q=75)
![Exercise 8.3
Hints: Getting Started | I'm Stuck
Suppose a 29.7 kg child sits 2.00 m to the left of center on the same seesaw. A
second child sits at the end on the opposite side, and the system is balanced.
(a) Find the mass of the second child.
mchild 2
kg
(b) Find the normal force acting at the pivot point.
Fn](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3a74cc31-62f4-4982-89c3-ce8d7d2cfbb2%2Ff3d657f1-cc11-4d90-9c7a-b94031bd7d78%2Fwudkmvs_processed.png&w=3840&q=75)
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