0.13 Phy1100: force -- vector solutions for coplanar forces concurrent  (Page 8/13)

Equal and opposite

The force vector pulling down on the ring has a magnitude of 10 newtons. We know because we tied a ten-newton load to the bottom of the ring. Therefore, themagnitude of the resultant vector pulling up on the ring must also be 10 newtons. That is the required length of the diagonal of the parallelogramdiscussed earlier.

The graphical solution

Draw the force vector pointing straight up from the ring with a length of ten units. (You can use whatever drawing scale is convenient for your drawing.) Thatis the diagonal of a parallelogram. Two force vectors lie along the lines of the two cords. Those two force vectors form the adjacent sides of a parallelogram,but we don't yet know their lengths.

Complete the parallelogram

The lengths of the force vectors that lie along the lines of the cords are equal to the sides of the parallelogram. That provides the answer to the originalquestion regarding the tension in each of the cords. The tension in each cord is equal to the length of the side of the parallelogram using the same scale thatwas used to draw the 10-newton diagonal length.

Drawing the parallelogram might be difficult

Drawing the parallelogram is not particularly easy even for a sighted person without special drawing tools. Here is one way that Ihave found to do it.

Draw a horizontal line (parallel to the pipe) that just touches the tip of the force vector that represents the diagonal. Then use your protractor to drawlines that begin at the tip of the force vector and emanate downward on both sides of the diagonal at angles of 60 degrees south of east and 45 degrees southof west. Be sure to do it in the correct order so that you end up with a parallelogram.

Mark the spots

Mark the spots where those lines cross the two upper cords. Those spots are the tips of the force vectors in the cords.

My answer

When I did it graphically (using pencil, paper, protractor, and ruler), I got the length of the force vector pointing 45 degrees north of east to have alength of 5.3 newtons. I got the length of the force vector pointing north of west to be 7.3 newtons. These are the values on the upper force vectors in theimage in Phy1100c1.svg.

Numeric sum exceeds 10 newtons

You may have noticed that the numeric sum of the magnitudes of these two force vectors exceeds the load of 10 newtons. How can that be?

The answer is that when force vectors that are supporting a vertical load go off in a direction other than vertical, only the vertical component of the forcevector counts insofar as supporting the load is concerned.

The vertical components of the two force vectors are:

• 5.3*sin(45 degrees) = 3.74
• 7.3*sin(60 degrees) = 6.32

The sum of the vertical components is:

6.32 + 3.74 = 10.06

which is close enough to the downward force being exerted on the ring by the mass.

Your answer

Were you able to get reasonably close to my answers? If so, congratulations. If not, don't worry too much about it. The most important thingis for you to understand the process and not to develop expertise in constructing accurate drawings using pushpins and rubber bands. However, I do believe that constructing vectordiagrams will help you gain a better understanding of the process.