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MindTrap
A horse travels a certain distance each day. Strangely enough, two of its legs travel 30 km each day and the other two travel 31 km each day. How is this possible?
TRANSLATIONALEQUILIBRIUM
TRANSLATIONALEQUILIBRIUM
In previous sections, something was always happening: forces were moving objects; things were colliding or energy was being transformed. Now we are going to study the “not-moving”, where things are “static” or when forces are in equilibrium (Latin for “equal forces”). Why are Statics important? The majority of objects and systems around you are at rest, suggesting that the forces on them are “in equilibrium”. For example, a book on a table has a force of gravity downward, but a force upward from the table (F
N) to
balance it and ‘cancel’ the gravitational force, keeping the book in equilibrium. Statics is concerned with the calculation of the forces acting on and within structures that are in equilibrium.
Ex 1.
A 20.0 kg object is suspended by a rope as shown. What is the net force acting on it?
Since the box is suspended, it is not accelerating so a = 0 m/s2
Fnet
= maF
net = 20(0)
Fnet
= 0 N
Ex 2.
Ok that was easy, now that same 20.0 kg object is lifted at a velocity of 4.9 m/s. What is the net force acting on it?
If the box is being lifted at a constant velocity, acceleration is still 0 m/s2 so F
net is still 0 N.
Because in both cases the net force on the objects is zero they are said to be in____________________.
An object is in equilibrium if its net force is equal to ________________ and therefore the object's acceleration must be ______________. Note that if an object is moving with constant velocity, it is still considered to be in translational (static) equilibrium.
Because in both cases the net force on the objects is zero they are said to be in equilibrium.
An object is in equilibrium if its net force is equal to zero and therefore the object's acceleration must be zero. Note that if an object is moving with constant velocity, it is still considered to be in translational (static) equilibrium.
Free Body Diagram Vector Diagram
Note that the vectors form a closed triangle;
the resultant of the three forces is zero!
F1
F2
F3
F3
F1
F2
● The block below will be in static equilibrium if the vector forces add to zero:
The triangle is closed because the resultant (net) vector is 0.
In this unit, all our vector diagrams will give us a closed triangle with no resultant so all vectors should be tip-to-tail
Translational motion refers to motion along a line, therefore:
The condition of equilibrium:
And so,
F⃗ net=Σ F⃗=0
F netx=Σ F x=0
F nety=Σ F y=0
Ex 3.
A 64 N object is suspended using ropes as shown in the diagram. Calculate tensions T1 and T2 in the ropes.
35o 50o
T1 T2
Strategy 1: Components1. Choose a point in the system that is in equilibrium, with all forces acting on it. In this case use the connection point of the ropes.
2. Draw an FBD!
3. Break these forces into x and y component. Note that we cannot arrive at numerical values for each component; we have to leave some as algebraic expressions.
4. Use the x and y Fnet equations to create a system of equations and solve them.
Strategy 1: Components1. Choose a point in the system that is in equilibrium, with all forces acting on it. In this case use the connection point of the ropes.
2. Draw an FBD!
3. Break these forces into x and y component. Note that we cannot arrive at numerical values for each component; we have to leave some as algebraic expressions.
4. Use the x and y Fnet equations to create a system of equations and solve them.
Although possible, this method is much more 'math heavy' than the second method.
2. Break each force into its x and y components using SOH CAH TOA1. Draw an FBD
cos35=T 1 x
T 1
→T 1 x=T 1cos 35
sin 35=T 1 y
T 1
→T 1 y=T 1 sin 35
cos50=T 2 x
T 2
→T 2x=T 2 cos50
sin 50=T 2 y
T 2
→T 2 y=T 2 sin 50
F gx=0NF gy=64N
3. Set upF netx and solve forT 1in terms of T 2
F netx=Σ F x0=F positive x−F negative x
0=T 2 x+F gx−T 1x
0=T 2 cos50+0−T 1 cos35T 1 cos35=T 2 cos50
T 1=T 2cos50cos 35
4. Set up F netyF netx=ΣF y
0=F positive y−F negative y0=T 1 y+T 2 y−F gy
0=T 1 sin 35+T 2sin 50−64
Continued on next slide
35° 50°
T1 T2
Fg
Ex 3.
A 64 N object is suspended using ropes as shown in the diagram. Calculate tensions T1 and T2 in the ropes.
35o 50o
T1 T2
Strategy 1: Components (continued)Strategy 1: Components (continued)
Although possible, this method is much more 'math heavy' than the second method.
0=T 1 sin 35+T 2sin 50−64
0=(T 2cos 50cos 35
)sin 35+T 2sin 50−64
64=T 2cos50sin 35
cos 35+T 2sin 50
64=T 2(cos50sin 35
cos35+sin 50)
64cos50sin 35
cos35+sin 50
=T 2
T 2=52.6260 ...→52.6 N
5. Substitute the expression found for T 1 instep 3 into the equation from step 4 andsolve for T 2
6. Substitute the value for T 2 found in step 5into the expression for T 1 found in step 3and solve for T 1
T 1=T 2cos50cos35
T 1=52.6260cos50cos 35
T 1=41.3 N
Strategy 2: Create a closed vector diagram1) Since we know that Fnet = 0 at any point in equilibrium,
what would happen if we added if we add up all of the force vectors?
They add to ZERO!!!
2) Use Sine Law, Cosine Law, or whatever means necessary to solve the triangle. NEVER assume that it is a right angle triangle unless you can prove it geometrically.
1. Draw an FBD
2. Add the vectors tip-to-tail to make a closed triangle.
35° 50°
T1 T2
Fg
35°
50°
T1
T2
Fg
3. Use trigonometry to find missing angles in the triangle. Notice that the dotted lines and F
g connect
at 90° because they are horizontal and F
g is vertical.
35°
50°
T1
T2
64 N
55°
35°
40°
85°
4. Use the Sine Law to find the values of T
1 and T
2.
sin 8564
=sin 40T 1
T 1=41.3 N
sin 8564
=sin 55T 2
T 2=52.6N
Ex 4.
An object is suspended as shown. If the tension in one of the ropes is 50 N as shown, what is the weight of the object?
37o
T1
T2 = 50 N
Ex 5.
Determine the weight, W, required to keep the system shown in static equilibrium.
