Statics: Equilibrium, Torque, and Applications - Lesson 3, Assignments of Physics

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LESSON 3: STATICS
Introduction
In this lesson you will study a special case in mechanics wherein the net force and the net
torque on an object or systems of objects are both zero. These are the two conditions for
equilibrium. Statics is concerned with the calculation of forces acting on and within structures
that are in equilibrium.
Learning Outcomes
After successful completion of this lesson, you should be able to:
Study the conditions that are necessary for an extended body to remain stationary, both in
terms of its position in space and its rotational ability.
 Express the conditions of equilibrium in the form of mathematical equations and apply them in
solving and analyzing variety of situational problems.
Discussion
3.1 First Condition of Equilibrium
If the vector sum of all the forces acting on the body is zero, then there will not be any change in
its state of motion, and the body is in a state of equilibrium under the action of these set of
forces.
It satisfies then that the sum of the x components (F1x, F2x, F3x . . . ) of all forces acting on the
object must be zero, and the sum of the y components (F1y, F2y, F3y, . . .) must also be zero.
These two equations are useful in calculating one or two unknown forces acting on an object if
all other forces acting on it are known. To use these equations correctly, here are some steps to
follow:
1. Make a drawing of the whole structure being considered.
2. Draw a separate force diagram known as the free body diagram for one part of the structure.
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LESSON 3: STATICS

Introduction In this lesson you will study a special case in mechanics wherein the net force and the net torque on an object or systems of objects are both zero. These are the two conditions for equilibrium. Statics is concerned with the calculation of forces acting on and within structures that are in equilibrium. Learning Outcomes After successful completion of this lesson, you should be able to:  Study the conditions that are necessary for an extended body to remain stationary, both in terms of its position in space and its rotational ability.  Express the conditions of equilibrium in the form of mathematical equations and apply them in solving and analyzing variety of situational problems. Discussion 3.1 First Condition of Equilibrium If the vector sum of all the forces acting on the body is zero, then there will not be any change in its state of motion, and the body is in a state of equilibrium under the action of these set of forces. It satisfies then that the sum of the x components (F1x, F2x, F3x... ) of all forces acting on the object must be zero, and the sum of the y components (F1y, F2y, F3y,.. .) must also be zero. These two equations are useful in calculating one or two unknown forces acting on an object if all other forces acting on it are known. To use these equations correctly, here are some steps to follow:

  1. Make a drawing of the whole structure being considered.
  2. Draw a separate force diagram known as the free body diagram for one part of the structure.
  1. Superimpose an x-axis and a y-axis on the force diagram.
  2. Write the equations for the first condition of equilibrium.
  3. Substitute the known information into the equations and algebraically solved for the unknowns. Sample Problems with Solutions :
  4. A 50-N body is supported by two ropes, one making 30 with the vertical and another making 45 with the horizontal as shown in Fig. 5.1. What is the tension in the ropes?

Determine the weights W1, and W2 shown in Fig. 5.3, that cause the tension T in the horizontal cable to be 64 N.

Torque is a measure of the tendency of a force to cause an object to rotate. The magnitude of torque (L) is the product of the magnitude of force (F) and the moment arm (s). The Moment Arm is the perpendicular distance from the line of action of the fore to the point about which the object rotates called axis of rotation. Point O is the axis of rotation, and S is the moment arm, the broken lines drawn from the force F is the line of action of the force. Hence, The torque is positive if the force tends to rotate the object counterclockwise about the origin and negative if it tends to rotate the object clockwise. Sample Problem:

Calculate the torque about point O caused by forces F1, F2 and F3. The magnitude of all the forces is 120 N and are applied a distance of 2.0 m from point O. 3. Second Condition of Equilibrium For an object to remain in rotational equilibrium, the positive torque that tends to rotate it counterclockwise must be balanced by a torque of equal magnitude that tends to rotate it clockwise. Thus, the second condition of equilibrium is:

  1. A uniform bar 3 m long is held by ropes at the ends making angle 60^0 and 30^0 , respectively, with the horizontal. A weight of 200 N is hung 0.5 m from the left end where the 60^0 rope is attached. Find the tension in the rope and the weight of the bar.

Watch the videos for further understanding of the lesson: Statics Lecture 01: What is statics? Statics: Crash Course Physics # https://www.youtube.com/watch?v=9cbF9A6eQNA Static Equilibrium - Tension, Torque, Lever, Beam, & Ladder Problem - Physics https://www.youtube.com/watch?v=qGvFAl5CK_c&t=814s

  1. Find the magnitude and direction of F3 in Fig. 2.7 for the given system to be in equilibrium.
  2. A uniform pole 20 ft. long and weighing 80 lbs is held by a boy 2 ft from one end while a man carries the same pole 5 ft from the other end. At what point should a load of 100 lbs be placed so that the man will carry twice as much weight as the boy.
  3. A springboard diver of weight 582 N stands at one end of a uniform 4.48 m diving board of weight 142 N. The board is attached to two pedestals, one at the other end of the board and the second pedestal 1.55 m from the first. Find the force in each of the two pedestals.
  4. A ladder, whose length is 10 m and whose mass is 40 kg rests against a frictionless vertical wall. Its upper end is a distance of 7.7 m above the ground. The center of mass of the ladder is of the way up the ladder. The coefficient of static friction between the ladder and the ground is 0.53. If a carpenter climbs 85% of the way up the ladder before it starts to slip, find the mass of the carpenter. 3 1