Translational Equilibrium Physics Montwood High School R. Casao.

Translational Equilibrium

Physics

Montwood High School

R. Casao

• Concurrent forces are those forces that are applied to or act on the same point, as shown at the point where T1, T2, and T3 meet.

• When two or more forces act at the same point, the resultant force is the sum of the forces applied at that point.

• The resultant force is the single force that has the same effect as the two or more forces that act together. • When forces act in the same direction or opposite directions, the total force can be found by adding the forces that act in one direction and subtracting the forces that act in the opposite direction.

• Equilibrium is the state of a body in which there is no change in its motion.

• A body is in equilibrium when the net force acting on it is zero; there is no acceleration (a = 0 m/s2), the body is either at rest or is moving at a constant velocity.

• The study of objects in equilibrium is called statics.• When two or more forces are acting at the same

point, the equilibrant force is the force that when applied at that same point produces equilibrium.

• The equilibrant force is equal in magnitude to that of the resultant force but acts in the opposite direction.

• If an object is in equilibrium in one dimension, the forces in one direction must equal the forces in the opposite direction.

ΣF+ direction = ΣF- direction

• If an object is in equilibrium in two dimensions, the net force acting on it must be zero.– For the net force to be zero, the sum of the x-

components must be zero and the sum of the y-components must be zero.

– Conditions for equilibrium:

• In general, to solve equilibrium problems:– Draw a free-body diagram from the point at which the

unknown forces act.

– Find the x- and y-components of each force.

– Substitute the components in the equations:

– Solve for the unknowns. This may involve

two simultaneous equations.

0

0x x x

y y y

A B C

A B C

0

0x

y

F

F

Two Conditions for Equilibrium• An object is said to be in equilibrium if and only if

there is no resultant force and no resultant torque (rotation).– First condition (translational equilibrium):

ΣFx = 0; ΣFy = 0

– The linear speed is not changing with time.

Car at rest Constant speed

0; F 0; No change in a v

– Second condition (rotational equilibrium)

Σtorque = 0; the object does not rotate, or rotates at a constant number of turns per unit time.

– Torque is the tendency of a force to rotate an object around an axis; torque measures how hard something is twisted.

Wheel at rest Constant rotation

0; . No change in rotation

Translational Equilibrium Only• If all forces act at the same point, then there is no

torque to consider and one need only apply the first condition for equilibrium: ΣFx = 0; ΣFy = 0

• Example: Find the tension in ropes A and B.

80 N

AB

600

80 N

A

B

600

Free-body Diagram:

By

Bx

• Read problem; draw sketch; construct a free-body diagram, indicating components.

• Choose x-axis horizontal and choose right direction as positive (+). There is no motion.

• The components Bx and By can be found from right triangle trigonometry.

y

x

R

2 2 2

sin ; sin

cos ; cos

yy R

Rx

x RR

R x y

Bx = B cos 600 By = B sin 600Bx = B cos 600 By = B sin 600

80 N

AB

600

80 N

A

B

600

Free-body Diagram:

By

Bx

• ΣFx = 0; Bx – A = 0; Bx = A

B·cos 60 = A; two unknowns

proceed to ΣFy = 0

• ΣFy = 0; By – 80 N = 0; By = 80 N; B·sin 60 = 80 N

• Determine A:

A = B·cos 60 = 92.38 N·cos 60 = 46.19 N

80 N

A

B

600

Free-body Diagram:

By

Bx

Fx = 0 Fy = 0

8092.38

sin60N

B N

Problem Solving Strategy

1. Draw a sketch and label all information.2. Draw a free-body diagram.3. Find components of all forces (+ and -).4. Apply First Condition for Equilibrium:

ΣFx = 0; ΣFy = 05. Solve for unknown forces or angles.

Example: Find Tension in Ropes A and B.

300 600

AB

400 N

AB

400 N

1. Draw free-body diagram.

2. Determine angles.

300 600300 600

Ay

By

Ax Bx

3. Draw/label components.

Next we will find components of each vector.

Next we will find components of each vector.

Example: Find the tension in ropes A and B.

Fx = Bx - Ax = 0

Fy = By + Ay - W = 0

Bx = Ax

By + Ay = W

AB

W 400 N

300 600Ay

By

Ax Bx

4. Apply 1st Condition for Equilibrium:

First Condition for Equilibrium:

Fx= 0 ; Fy= 0

Example: Find the tension in ropes A and B.

Bx = Ax

By + Ay = W

Using Trigonometry, the first condition yields:

B cos 600 = A cos 300

A sin 300 + B sin 600 = 400 N

Ax = A cos 300; Ay = A sin 300

Bx = B cos 600

By = B sin 600

Wx = 0; Wy = -400 N

Example: Find the tension in A and B.

cos301.73

cos60

AB A B = 1.732· AB = 1.732· A

We will first solve the horizontal equation for B in terms of the unknown A:

B cos 600 = A cos 300

A sin 300 + B sin 600 = 400 N

We now solve for A and B: Two Equations and Two Unknowns.

Example: Find Tensions in A and B.

A·sin 300 + B·sin 600 = 400 NB = 1.732·A

A·sin 300 + (1.732 A)·sin 600 = 400 N

0.500·A + 1.50·A = 400 N2·A = 400 N A = 200 NA = 200 N

B = B = 1.7321.732 A A

Now apply Trig to:

Ay + By = 400 N

A sin 300 + B sin 600 = 400 N

Example: Find B with A = 200 N.

Rope tensions are: A = 200 NB = 346 N

B = 1.732·A

A = 200 N

B = 1.732·(200 N)

B = 346 N

AB

W 400 N

300 600Ay

By

Ax Bx

• You may need to find the tension or compression in part of a structure, such as in a beam or a cable.

• Tension is a stretching force produced by forces pulling outward on the ends of an object.

• Compression is a force produced by forces pushing inward on the ends of an object.