MOMENTS IN STATIC SYSTEMS: Free-Body Diagram of Boom (Figure. 2) **PLEASE THOUROUGHLY ANSWER THE THREE QUESTIONS BELOW**
Posted: Thu Feb 17, 2022 11:05 am
MOMENTS IN STATIC SYSTEMS: Free-Body Diagram of Boom (Figure.
2)
**PLEASE THOUROUGHLY ANSWER THE THREE QUESTIONS
BELOW**
Purpose: The purpose of this experiment is to determine the
effects of various forces and torques acting upon a rigid body in
equilibrium.
THEORY
When a body is acted upon by
concurrent, coplanar forces, the body is in equilibrium when the
vector sum of the applied forces is zero. This was demonstrated in
Experiment M-2.
If, however, the forces are
nonconcurrent -- that is, they all do not act through the same
point -- additional conditions must be established if equilibrium
is to exist. These forces must not produce any angular
acceleration. Since angular acceleration is produced by the action
of torques (or moments of force), the algebraic sum of these
torques about any axis must be zero if the body is to remain in
equilibrium. The torque for each force with respect to the selected
axis is defined as the magnitude of the force times the
perpendicular distance from the axis to the line of action of the
force. Counterclockwise torques are assumed to be positive and
clockwise torques negative.
Figure 2 is a free-body
diagram of the boom. This sort of diagram is necessary if
we are to determine analytically the effect of the various forces
and torques on the boom. The boom is in equilibrium. Since it is
not accelerating in the x direction, equation (l) must hold true;
since it is not accelerating in the y direction, equation (2) must
hold true; and since it has no rotational acceleration, equation
(3) must hold true. The torque equation is written for the axis at
the pin O at the lower end of the boom.
(Equation. 1)
βππ = π
π― β S
= π
(Equation. 2)
βππ = π
π½ β π³ β ππ =
π
(Equation 3.)
βΞπ = π
πΆπͺ Γ πΊ β πΆπ¬ Γ
ππ β πΆπ« Γ π³ = π
QUESTIONS
1. Suppose that
you were unaware of the balance of moments requirement for
equilibrium (SG = 0) and that the only requirements known to you
were the balance of forces (SFx = 0, SFy = 0). In Figure 2 would it
be possible for you to calculate the force S, knowing only the
weights L and w and the various distances? How many unknowns and
how many independent equations would you have? Is it possible to
make this calculation with the additional equation for the balance
of moments?
2. Any two forces
acting on a body can be combined into a single resultant force
having the same effect. Is this statement always true?
Explain.
3. If the
summation of moments had been made about point D would the results
have been different? Explain.
Figure 2 Free-Body Diagram of Boor S L 0 HE D
2)
**PLEASE THOUROUGHLY ANSWER THE THREE QUESTIONS
BELOW**
Purpose: The purpose of this experiment is to determine the
effects of various forces and torques acting upon a rigid body in
equilibrium.
THEORY
When a body is acted upon by
concurrent, coplanar forces, the body is in equilibrium when the
vector sum of the applied forces is zero. This was demonstrated in
Experiment M-2.
If, however, the forces are
nonconcurrent -- that is, they all do not act through the same
point -- additional conditions must be established if equilibrium
is to exist. These forces must not produce any angular
acceleration. Since angular acceleration is produced by the action
of torques (or moments of force), the algebraic sum of these
torques about any axis must be zero if the body is to remain in
equilibrium. The torque for each force with respect to the selected
axis is defined as the magnitude of the force times the
perpendicular distance from the axis to the line of action of the
force. Counterclockwise torques are assumed to be positive and
clockwise torques negative.
Figure 2 is a free-body
diagram of the boom. This sort of diagram is necessary if
we are to determine analytically the effect of the various forces
and torques on the boom. The boom is in equilibrium. Since it is
not accelerating in the x direction, equation (l) must hold true;
since it is not accelerating in the y direction, equation (2) must
hold true; and since it has no rotational acceleration, equation
(3) must hold true. The torque equation is written for the axis at
the pin O at the lower end of the boom.
(Equation. 1)
βππ = π
π― β S
= π
(Equation. 2)
βππ = π
π½ β π³ β ππ =
π
(Equation 3.)
βΞπ = π
πΆπͺ Γ πΊ β πΆπ¬ Γ
ππ β πΆπ« Γ π³ = π
QUESTIONS
1. Suppose that
you were unaware of the balance of moments requirement for
equilibrium (SG = 0) and that the only requirements known to you
were the balance of forces (SFx = 0, SFy = 0). In Figure 2 would it
be possible for you to calculate the force S, knowing only the
weights L and w and the various distances? How many unknowns and
how many independent equations would you have? Is it possible to
make this calculation with the additional equation for the balance
of moments?
2. Any two forces
acting on a body can be combined into a single resultant force
having the same effect. Is this statement always true?
Explain.
3. If the
summation of moments had been made about point D would the results
have been different? Explain.
Figure 2 Free-Body Diagram of Boor S L 0 HE D