0.15 Phy1120: force -- center of gravity  (Page 7/8)

The lines K1 , K2, Z1, Z2, and R

To answer this question, begin by drawing a line from the C.G. to point b for each box. Label this line K1 for the left box and K2 for the right box.

Label the line segment that extends from the C.G. to the bottom of the left box as Z1 and label the line segment that extends from the C.G. to the bottom ofthe right box as Z2. Label the line segment that is the right half of the bottom of each box as R.

Right triangles

In the left box, the line segments K1, Z1, and R form a right triangle for which the lengths of two line segments are:

• Z1 = 1
• R = 1

In the right box, the line segments K2, Z2, and R form a right triangle for which the lengths of two line segments are:

• Z2 = 4
• R = 1

Compute the lengths of K1 and K2

Using the Pythagorean theorem and the Google calculator, we can compute the following lengths:

K1 = sqrt(1^2 + 1^2) = 1.414

K2 = sqrt(1^2 + 4^2) = 4.123

The sides of the right triangles

Next, we will compute the interior angle for each triangle at the C.G. vertex. Label the interior angle for the left box P1 and label the interiorangle for the right box P2.

We know the lengths of all three sides of each triangle. There are a couple of ways that we can use trigonometry to find the interior angles. Considering thetriangle from the viewpoint of the interior angle at the C.G. for the left triangle:

• hypotenuse = K1 = 1.414
• base = Z1 = 1
• opposite side = R = 1

For the right triangle,

• hypotenuse = K2 = 4.123
• base = Z2 = 4
• opposite side = R = 1

Compute the interior angles at each C.G. vertex

Using the Google calculator, we can compute the interior angle at the C.G. for each box as follows:

• P1 = arcsine(1/1.414)*180/PI = 45 degrees
• P2 = arcsine(1/4.123)*180/PI = 14 degrees

Tilt each box and observe the angles at the C.G. vertices

Now pretend that you exert a force on the top-left corner of each box causing it to rotate around the lower-right corner at point b. As you do that,the weight vector would rotate around the C.G. in a counter-clockwise direction (always pointing straight down) and the angle between the weight vector and theline K1 or K2 would decrease.

The tipping point

When the angle between the weight vector and the line K1 or K2 goes to zero degrees for either box, the tipping point for that box will havebeen reached. (See the image in the file named Phy1120a1.svg, or Figure 1 .)

At that exact point, the box will still be in equilibrium, but it will be in unstable equilibrium. Exerting a little more force to the right will cause the box to turn all the wayover onto its side under its own weight. Relaxing the force a little when the box is in that position will allow it to rotateback towards its original position.

What are the tipping point angles?