Mechanical equilibrium is where there is no acceleration, meaning there are no changes in translational velocity ([latex]\sum{F}=0[/latex]), known as translational equilibrium; and no changes in rotational velocity ([latex]\sum{M_{O}}=0[/latex]), known as rotational equilibrium. This will mean that the [latex]\sum{F_{upward}}=\sum{F_{downward}}[/latex], and [latex]\sum{F_{left}}=\sum{F_{right}}[/latex]; and [latex]M_{clockwise}=M_{anticlockwise}[/latex]. This does not mean it has to be stationary; it can be moving, but it must be moving with constant velocity. This translates to meaning the sum of forces on all particles is [latex]0[/latex], and sum of all moments on all particles is [latex]0[/latex]. Moment [or torque] is the tendency of a force to rotate, and is [latex]M_{O}=r\times F=r.F.sin(\theta)[/latex], and has the units [latex]Nm[/latex]. The lever arm ([latex]r.sin(\theta)[/latex]) is the distance from the point of rotation, to where the force intersects at right angles (which is what cross product, or [latex]sin(\theta)[/latex] achieves) with the lever arm. Calculation of these questions require drawing a free body diagram, which is a diagram depicting all forces and moments, and ensuring all forces and moments net [latex]0[/latex]. For example, in the diagram , there are no moments. The system chosen (particularly in determining moment, as moment is generated when there is force not acting on the point chosen), should be where the ropes attach, as there is no moment (spin) caused by forces at this point. Equating the forces leftward and rightward, [latex]a.cos(45^{\circ})=b.cos(45^{\circ})[/latex], meaning [latex]a=b[/latex]. Force downward can be found with [latex]F=mg=5\times 10=50[/latex]. Equating the forces upward and downward, [latex]a.sin(45^{\circ})+b.sin(45^{\circ})=50[/latex], meaning [latex]a+b=\dfrac{50}{sin(45^{\circ})}[/latex]. Since [latex]a=b[/latex], [latex]2a=\dfrac{50}{sin(45^{\circ})}[/latex], meaning [latex]a=\dfrac{25}{sin(45^{\circ})}=35[/latex]. Thus, both [latex]a=b=35N[/latex].

[img]translational-equilibrium-example.png[/img]

Another example, where there are two weights, 50kg and 10kg, how far away from either mass does the (massless) stick need to be held to achieve balance? Since [latex]F_{upward}=F_{downward}[/latex], [latex]F_{upward}=50(10)+10(10)=600N[/latex], which can only be caused by the finger holding the stick up. Also, [latex]M_{clockwise}=M_{anticlockwise}[/latex]. The point chosen should be where the most forces pass through, for example, where the [latex]50kg[/latex] weight is hanging. This means that the moment at the point at which the stick is held, is [latex]Fd=600\times anticlockwise[/latex]. The other moment due to the 10kg weight is [latex]Fd=10(10)\times 1=100clockwise[/latex]. Thus, equating moments clockwise and anticlockwise, [latex]100=600x[/latex], meaning [latex]x=\dfrac{1}{6}=0.17[/latex], meaning that the stick should be held 17cm from the left end to balance the masses. If the stick isn’t massless, the weight of the stick is visualized at its center of mass, which can be used to determine its according force and torque, to also consider in the equilibrium equations.

[img]rotational-equilibrium-example.png[/img]

Frequently asked questions

What is equilibrium? Just as the plain word suggests, it means things are in balance.

What is translational equilibrium? How does that differ from rotational equilibrium? Translational equilibrium is where translational forces left-and-right, and up-and-down, are in balance. Rotational equilibrium is where moments clockwise-and-anticlockwise are in balance.

How can there be equilibrium if there velocity? It makes sense stationary, but... moving?? Because translational equilibrium is where forces are balanced. And force is [latex]F=ma[/latex]. So therefore, force doesn't change if acceleration doesn't. And when velocity is constant, acceleration doesn't [change].

In contrast, non-equilibrium is where there is acceleration. Since [latex]F=ma[/latex], these questions can be solved with [latex]\sum{F}=ma[/latex]. For example, a 5kg mass hanging from a rope accelerates at [latex]5m/s^2[/latex]. Assigning down as positive, gravity pulls down at [latex]F=mg=5\times 10=50N[/latex]. Tension is negative as it is in the opposite direction. Thus, [latex]50-tension=ma=5\times 5=25N[/latex]. Thus, [latex]tension=25N downwards[/latex]. If the variable found is negative, it means that the direction is opposite to the one set.

Soon, Jamie’s relationship with the blonde klutz moved from mere friends to, well, hand holding .

“For Christians, that’s an engagement,” Mandy giggled.

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Maps to RK3.A

What is equilibrium?

2 Momentum

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Whereas inertia was the tendency of an object to remain in its present state of motion, momentum is the tendency to continue moving in its present direction. Remember inertia is measured by mass, and is hence an innate property of an object, and doesn’t change. In contrast, momentum can change. Momentum is defined as [latex]p=m.v[/latex], meaning momentum is directly proportional to mass and velocity. Momentum has the SI units [latex]kg m/s[/latex], and is a vector quantity. Due to Newton’s 1st law, a force is necessary to stop motion. This force is greater when either mass of the object or velocity is greater. As a result, a greater momentum indicates increased difficulty to stop, and a lesser momentum indicates increased ease to stop. Where there is conservation of both mass and energy (known as a closed system, discussed ), total momentum is conserved.

So that was how Jamie came to know Sophie.

“That story was… alright,” Mandy started, “but it’s like a little spark compared to how WE met.”

Mandy was an engineering student at Pacific Coast College, and after noticing Jamie in the library on several occasions, felt led to talk to him.

“Hey, can I use your back to draw on?” Mandy asked.

“I still remember that white wide headband you wore that day,” Jamie commented, “so preppy LOL.”

“Umm,” Jamie replied, bedazzled, thinking what sort of stranger would want to do that, and what Sophie would think, before, being polite, answered, “I guess so.”

“Can you lean forward?” Mandy asked.

Mandy didn’t say much whilst she was filling out the class questionnaire on Jamie’s back. In reality, she was thinking how she’d make acquaintance with Jamie.

“Okay, I’m done,” Mandy remarked, as Jamie turned back around, “the name’s Mandy, I study engineering.”

“I’m Jamie, I study law,” Jamie replied.

“Umm, I think your zip is undone,” Jamie said to Mandy.

“Oh,” Mandy said zipping up, somewhat embarrassed, “uh… why were you looking down there LOL?!!”

“Slut alert,” Blaire commented, rolling her eyes.

Collision is the exertion of two forces on each other, and can be either:

Elastic collision, in which mechanical energy is conserved. Elastic collision is often said to conserve kinetic energy, but keep in mind mechanical energy is defined as kinetic and potential energy (see ). This means the statement is not always correct. For example, a ball dropped which compresses a string, converts all kinetic energy into elastic potential energy, thereby still conserving all mechanical energy (despite there is no more kinetic energy). Remember conservation of mechanical energy can only be caused by conservative forces (see ). For example, a bouncy ball dropped from a height, bouncing back up to the same height, starts with gravitational potential energy and finishes with the same energy, and therefore collides elastically

Inelastic collision, in which mechanical energy is not entirely conserved. Some mechanical energy is instead converted into heat. Remember conservative forces will lead to conservation of mechanical energy (see ), meaning inelastic collisions require at least one non-conservative force. Generally, when there is increase in temperature, or deformation (though see exception for conversion into elastic potential energy, see ) of one or more of the objects, the collision will be inelastic. For example, an object dropped which shatters, starts with gravitational potential energy, but finishes with no potential energy, and therefore collides in-elastically

In both elastic and inelastic collisions, momentum is conserved. As momentum is a vector, [latex]p_{initial (vertical)}=p_{final (vertical)}[/latex] can be resolved into components, such that [latex]p_{initial (vertical)}=p_{final (vertical)}[/latex] and [latex]p_{initial (horizontal)}=p_{final (horizontal)}[/latex]. Collision problems can be solved by drawing a diagram of before the collision, and after the collision. Evidently, if the collision is elastic, initial and final energy (which is a scalar, see ) should be set to equate each other, as no vectors are required, and therefore no diagrams are required.

Reverse collision (aka explosion) are derivatives of collisions, which rather than coming together, start together and separate. For example, an astronaut who throws a rock in space. Throwing the rock in one direction will causes an equal and opposite momentum on the astronaut in the other direction. Reverse collision problems can be solved by noting that [latex]p_{initial}=0[/latex] as objects are initially stationary.

Impulse shows that a force sustained over a period of time produces a change in momentum, more than a force applied briefly. Impulse is change in momentum, and is defined as [latex]I=\Delta mv=F.\Delta t[/latex]. This can be derived from [latex]F=ma[/latex], since [latex]a=\dfrac{\Delta v}{t}[/latex], so substituting in, [latex]F=\dfrac{m.\Delta v}{t}[/latex], or reshuffling, [latex]F.t=m.\Delta v[/latex]. Impulse is the integral of force with respect to time. Integral is the opposite of differentiation. Differentiation is the slope of a line. Integral thus finds the area under the graph of a function, in the case of impulse, when force is graphed against time.

Formative learning activity

Maps to RK3.B

What is momentum?

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