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posted Jun 18, 2016, 4:51 PM by Upali Salpadoru   [ updated Jul 17, 2016, 1:31 PM ]
Fig.1 John using a lever.

John is moving a heavy stone with a strong steel bar. When he pushes down, the load goes up. When you use a rod in this manner to change  a force in direction or size (magnitude) , we call it a LEVER.

There are three different ways of using levers. It all depends on the order of using Effort, Pivot or Fulcrum, and Load.

Class 1           EFFORT == PIVOT == LOAD  or    LOAD ==PIVOT == EFFORT.

Class 2.         EFFORT ==LOAD == PIVOT.

Class 3.         PIVOT==EFFORT == LOAD.

The three types are illustrated here.

Fig. 2 Three types of levers.


Lever problems can easily be solved by finding the Moments / Torque of a force.

One may think that  a lever can be balanced by making the forces turning clock wise equal to forces turning anti clock-wise. It is not so. You may have already seen a small kid lifting a bigger child in a seesaw.

What becomes equal are the moments. What is a moment?

A Moment or a Torque is the turning effect due to a force. This is obtained by multiplying the force by the distance to the pivot.

                      Moment = Force x  Distance to Pivot.

Law of Moments
If a Lever is balanced the sum of the clock-wise moments is equal to the sum of the anti-clockwise moments.

Find the effort required to lift the load.
Cw Moments = ACW moments.
 E x 10 + 760 x 0.2 = 200 x 1
          10 E + 15.2  =200
                    Effort =(200 - 15.2) /10

When the Pivot is not given?
 A log is supported by two pillers A and B. A Bunny telysits on it closer to B. It is clear that B gets most of the weight of Bunny. We have to find the support force at A and B separately.  Now where is the Pivot?
Fig.4 Bunny on a uniform log.
When the Pivot is not given as in this case , you can assume it to be anywhere and take moments around it.

Let us assume  P to be at A.
As the log is uniform, the centre of mass will be at the mid point of 5m. log.
The the CW moments are  =  (100x 2.5)  +  (40 x 4)
           ACW  moments      =  B x 5
  Therefore                5 B   =   250 + 160
                                  B    =   410/5  = 82 N.
We can make use of the parallel forces to find the force at A.
 Up forces = A + 82
 Down forces =  100 + 40
 Therefore  A + 82 = 140
                      A    =  140 - 82  =      58 N.      
Torques and Couples.

     How do you turn?

     Clockwise turn                               

Fig.5 Turning a car.                                    Anti clockwise turn.


 Have you  ever wondered what makes an object turn or twist?

For any kind of turn, there should be two forces acting in opposite directions. In physics, a combination of two such forces is called a  ‘couple’[1] .

We see the man standing on the plank. So his weight is acting down. There does not seem to be another force, but there is another force.

Fig. 6.  Where is the other force?

It is really the fixed point that gives the other force. This is really the reaction force as shown  here .

There are two directions of turn in any particular plane. Normally we consider the two important planes, the vertical plane and the horizontal plane. The sea-saw goes up and down in a vertical plane, while a merry-go-round turns in a horizontal; plane.

 Turning normally takes place around a point, which does not go up and down. In physics, we use three names for this. They are:  Fixed point -- Fulcrum - Pivot.

All three have almost the same meaning. In ordinary language words such as ‘hinge’ or ‘swivel’ are also used

A Couple is  usually defined as two Equal, Parallel and opposite  forces, not in the same line.

   Torque of a couple = A Single Force x Diameter.


The uniform board is pivoted at the mid point.Bunny standing 2.5m away from pivot balances 
Lily who is 2m from Pivot.
1. Find the CW moments.
2. Find the ACW moments. (Use lily's weight as 'L'.)
3  Find the weight of Lily.
4. Find the support/reaction force.
                                                        5x4 = 20 marks

 Q. 2.0

 1. Give the CW moments.

  2. Give the ACW moments.

  3. What will be the result due to the forces?

  4. If the 600 N force is made to act perpendicular to the lever, what would be the ACW moment?

  4. According to 4 what would be the result?

     5 4 = 20 marks.


Q.Q. 3.0


Load distance  =  40 cm

Effort distance = 5 cm.

Load                 = 25 N

Ignore the weight of the hand.

Consider the effort to be acting vertically.

Find the following:-

  1. Clockwise moment.

  2. Anticlockwise moment.

  3. Vertical force  to be exerted to lift the load.

  4. As the muscle is at an angle real force necessary would have to be higher or lower?

  5. If you consider the weight of the forearm, will it be helping you to lift?

  6. What class of a lever is this?

                                                           5 x 6= 30 Marks.

 Q. 4.0

LQ 5.jpg

  1. Taking A as the pivot find the CW moments.

  2. Then what are the ACW moments?

  3. What is the weight Ali is getting?

  4. What is the weight Bunty is getting?

  5. If Lizz moves closer to G how will the weights on Bunny and Ali will change? 5x5 - 25 Marks.

      Q 5.0

       LQ 5.jpgExplain how a couple is formed for the tyre to roll down the incline.


   5 Marks.

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