Equilibrium WorkbookName:Block:

  1. Equilibrium occurs when an object does not accelerate. This means that the object must have ______velocity. Often that velocity is zero. There are two ways that an object can accelerate.
  2. Longitudinal acceleration occurs when the object accelerates along a line. When an object is not accelerating along a line it is said to have translational equilibrium
  3. Rotational acceleration occurs when an object is made to rotate faster or rotate slower. When an object is not accelerating in this manner it is said to have rotational equilibrium.

Notice that for each of the pictures you drew the forces had to be balanced so that object would not accelerate. The equations for these two situations are shown below.

Translational Equilibrium

F= 0or Fright= Fleft& Fup= Fdown

Rotational Equilibrium

= 0or ccw= cw

We will look at each of these separately.

Translational Equilibrium

F= 0or Fright= Fleft& Fup= Fdown

  1. For translational equilibrium to occur the forces in the vertical and horizontal direction on an object must balance. To achieve translational equilibrium for a situation the steps to follow are:
  2. Resolve the existing forces into horizontal and vertical components (orthogonal components along x and y)
  3. Find a vector that will balance the forces in both dimensions.
  1. Resolve each of the following into orthogonal components along x and y.
  1. For the following diagrams a) Resolve each of the force vectors into x and y components. b) Use the stack method of working with vectors to determine the x and y components of the vector that would create translational equilibrium. c) Find the magnitude and direction of the vector in part b. d) Sketch the vector onto the diagram.
  1. For each of the diagrams, a 37 kg lamp hangs form the ceiling near a wall and a cable is used to hold it in place. Find the tension in the cable.

  1. A 60kg crate is suspended from ropes as shown. Calculate the tensile force exerted by each of the ropes.
  1. Do Equilibrium 1 Worksheet

Rotational Equilibrium

= 0or ccw= cw

  1. Rotational equilibrium occurs when the forces on an object are balanced so that the object cannot rotate. For an object to rotate it must rotate about a point – like a wheel rotates around the axel. The point of rotation is the center of the axel. The point of rotation is called a fulcrum. A force that causes an object to rotate acts on the object at a distance from the fulcrum and in a direction that is perpendicular to the radius of a circle from the fulcrum to the point of attachment of the applied force – see diagram.
  1. For each forces acting on the object below, determine if the force will cause rotation and the direction (clockwise or counter-clockwise) of the rotation and an expression for the component of that force that causes rotation.
  1. Why do people use screwdrivers and wrenches?
  2. The longer the wrench you use the greater the radius and the greater the torque. Torque is the value that makes something turn. A small force can act with a large radius to create significant torque. Your teacher will ask for a volunteer to hold a 1 kg mass using a meter stick. What do you notice about the amount of strength needed to hold the 1kg mass as it is moved along the length of the stick?

The symbol for torque is . For an object to remain in rotational equilibrium it must experience a net torque of zero. In other words, the torque that causes it to turn clockwise must be equal to the torque that causes it to turn counterclockwise. These concepts are summarized as follows.

= 0or ccw= cw

= F┴d

  1. For each of the vectors, calculate the magnitude of the torque, sketch in the direction of the rotation and label the torque as clockwise or counterclockwise.
  1. Two boys play on a seesaw. One of them weighs 48 kg the other weighs 41kg. If the seesaw is 3m long and balanced in the middle, then at what distance should the 48kg boy sit if the 41kg boy sits at a point 1.4m from the center? Solve this question by: a) Drawing a force diagram. b) labeling the direction of the torque produced by each boy. c) Writing the equation for the rotational equilibrium. d) solve.
  1. In the following diagram the mass of the horizontal beam is 5kg. Find mass, M, such that the beam is in equilibrium. (cg = center of gravity)
  1. In the following diagram, find the tension in the string. The beam is 2m long and has a mass of 25kg.
  1. A mass is suspended from a 2m, 500g, pole as shown and the pole is held up at each end by a bird. What is the upward force that each bird applies to the pole? To solve this problem choose the fulcrum to be at one of the birds and find the upward force imparted by the other bird. Then repeat using the opposite side as the fulcrum.
  1. Do #2, 4, 5, 7, 9, 10, 11, 12, 14 pp.230-231
  1. Do Activity “Equilibrium of a Loaded Beam”
  1. Cranes and Booms
  1. A 10m, 65kg boom extends horizontally from a wall and has a 50kg mass hanging from it 7.5m from the wall.

i)Find the tension in the cable.

ii)Find the vertical force at the wall.

iii)Find the horizontal force at the wall.

iv)The vertical and horizontal forces at the wall can be combined to find the reaction force at the wall. Find it.

  1. Find the reaction force at the wall for the following beam. The mass of the boom is 200kg and it is 15m long. The 75 kg mass is hanging off the far end of the boom

c)Find the tension and the reaction force at the wall for the following 12m, 125kg boom.


8. Ladders and Torque (To Slip or not to Slip, That is the Question)

If a ladder slips it's bottom slides away from the wall and it rotates as it falls. So, a ladder problem is a combination of Translational Equilibrium and Rotational Equilibrium. For the ladder not to slip,

 = 0or ccw = cw

and

F = 0or Fright = Fleft& Fup = Fdown

If the above conditions are not true then the ladder falls.

To simplify the problem a little we will assume that there is no friction between the ladder and the wall. This means that the wall only pushes horizontally on the top of the ladder and that there is no friction with which the wall can "help" hold the ladder up. While this might seem like a foolish assumption, it sure makes the problem easier to solve. Also, this is as hard as it gets for Phy12.

In the following problem you are going to find the shallowest possible angle for  in the diagram at the right.

For the ladder diagram at the right draw in the horizontal and vertical forces that are acting on the ladder. Use the following data:

Mass ladder = 10 kg

Length ladder = 3m

Mass paint bucket = 3.5 kg

 = 0.35

distance top of ladder to paint bucket = 0.75m

If all went well you should have five arrows on the diagram: the wall pushing perpendicularly away from itself, the floor pushing up, the force of friction pushing the ladder toward the wall at the floor, the weight of the ladder acting at its center, and the weight of the paint bucket.

Now using the arrows on the diagram and the information above, set up the equations for:

Fright= Fleft& Fup= Fdown

Next: Chose a fulcrum for the torque portion of this question. Usually the bottom is easiest to work with. Then set up all torques for: [Make sure your sine and cosines are correct! Combine the translational forceswith the torque. ]

ccw = cw

Solve for .

Practice Problems:

1)A 12kg ladder 5m long at an angle of 70degrees to the horizontal (the floor) starts to slide when a 65kg girl reaches a point 4m from the bottom. Calculate the coefficient of friction between the floor and the ladder.

2)A 10 kg, 3.0m long ladder at 60 degrees to the horizontal starts to slide when a 75kg man is too high up! If  = 0.4 between the ladder and the floor, then how high was he when the ladder started to slide?

3)For question two, what angle should the ladder be at so that the man can stand on the last wrung only 15cm from the top of the ladder.

9)Do worksheet “Equilibrium 2”

10)Do worksheet “Equilibrium 3”

TolksdorffEquilibrium Workbook1 of 12