Where do you place the pivot?

Question

 You are given a 2 m plank with a uniform mass of 10 kg. On one end of the plank, a 25 kg mass is being placed, i.e. the center of the mass is assumed to be at the end of the plank. On the other side a mass of 35 kg is being placed at exactly 0.25 meters from the end. You are in charge of setting a pivot such that the pivot is located at the balance point. Where do you place the pivot?

Answer 1

This sketch might help:

Explanation:

I arbitrarily assigned the 25 kg mass on the left end with the distance from the left end to the pivot point as "x".

The mass of the plank is 10 kg per 2 meters which means it is 5 kg/m. The amount of mass of the plank on the left side of the pivot point is the mass per unit length multiplied by the length = (5)(x) = 5x. The distance the center of mass is from the pivot point is x/2.

The mass of the plank on the right side of the pivot point is (5)(2 - x) and the distance the center of mass is from the pivot point is (2 - x)/2.

Set up an equilibrium equation where CCW torques are positive and CW torques are negative like this:

(25)(x) + (5x)(x/2) - (5(2-x))((2-x)/2) - (1.75 - x)(35) = 0 (using the pivot point as the "zero" location.

and solve for "x"

Answer 2

Hi Annet,

First, draw the system out (you would be surprised at how helpful this can be).

Also, if the plank is uniform that is code for the center of mass is located at the center of the plank.

Now you have three masses (each of which creates a torque due to their weight).

The key thing (in this case) is that the system is balanced.

You can therefore choose any point to represent x = 0 (imagine a number line running under the plank).

Edit:

Choose the left side of the plank to be zero (it's just easier).

Also, sum your forces to 0 since the system is in equilibrium.

Treat the mass of the plank as if it is located at the center, there is not need to find the linear density.

Let the pivot be located at x.

Be sure to draw all of the forces acting on the plank (including the normal from the pivot).

You will end up with one equation one unknown.

If you place the 35kg mass at x = 0.25m and the 25kg mass at x = 2m you should find the pivot to be located at x = (68.75/70) m.

Hope this helps!