Equilibrium of forces

Question

 A uniform meter rule of mass 90g is pivoted at the 40cm mark.if the rule is in equilibrium with an unknown mass M placed at a the 10cm mark and a 72g mass at the 70cm mark,determine M

Answer 1

Hi Nanyak!

 

Both of the problems you posted deal with the equilibrium of torques.  Torque is, informally, a measure of the effectiveness of a force in causing rotation, and is the product of the force, the magnitude of the displacement vector drawn from the rotation point to where the force is applied, and the angle between the force and that displacement "lever arm" vector.

 

In this case, all the forces are being applied to the meter rule (meter stick).  The distances from the forces to the pivot point are just the distances along the ruler, measured by their markings.  For example, the 72-g mass above is exerting a force on the meter stick at 70 cm, with the meter stick's pivot at 40 cm, so that displacement from the pivot to where that force is being applied is 30 cm.

 

So you have three forces providing torque to the meter stick:  the weight of the unknown mass M pulling down on it, the weight of the 72-g mass pulling down on it, and its own weight pulling down on it.  For calculating torque, you apply the object's own weight (in this case, the meter stick's own weight) at its center of mass, which -- since the meter stick is uniform -- should be in the center of the meter stick (at 50 cm).  

 

All the forces are applied down on the meter stick, and the displacement vectors for each force are drawn parallel to the stick, which is presumed to be oriented in balance, horizontal.  So the angle between the force and displacement vector is each case is 90 degrees, and sin(90) = 1.  Thus, the torque for each force in this configuration reduces to Fr, the magnitude of the force F times the magnitude of displacement r.

 

You also have to put an appropriate sign on each torque.  Torque causes objects to rotate, and -- for most of the cases we deal with at this level -- there are only two possible modes of rotation:  clockwise and counterclockwise.  Since there are only two directions, we can label them "+" or "-".  If we call counterclockwise "+," as is conventional, then any torque that makes the meter stick rotate clockwise will be "-," and those that make it rotate counterclockwise will be "+."  Clockwise and counterclockwise depend on whether you are looking from in front or behind, but once you choose a direction, you can consistently apply signs to each force, and you will get the same result (if you are consistent) no matter which side you look from... since nothing changes about the physical actions of the system.

 

So, for example, this 72-g force is just the weight (in Newtons) of the 72-g mass.  It is applied 30 cm (0.3 m) from the pivot, and -- when viewed from the "front," so that the 72-g mass is to the right of the pivot -- would make the stick rotate clockwise, or "-".  So the torque it creates around the pivot would be:

 

τ₇₂ = -(W₇₂)(0.03 m)

 

where W₇₂ is the weight (in newtons) of the 72-g mass.  You can get values and expressions for all the forces causing torques in this way, including the weight of the meter stick, which should be applied at the halfway (50 cm) mark.  If the stick is in equilibrium, all those torques should add up to zero.  Getting expressions for all the torques, adding them together, and setting them equal to zero should give you one equation with one unknown, the mass M you are asked for.  Then you can solve for M.

 

See if that gets you going, and if you get stuck, or want to check an answer, just let me know.