Chapter 9: Force

9.4: Problem solving steps

We’ll follow the same problem-solving steps as with problems dealing with conservation laws, but this time we’ll add an intermediate step between the drawing and the math:

  1. Draw a picture.
    • Using your picture, draw a free body diagram that represents the forces acting on one particular object.
  2. Using your free body diagram, apply Newton’s second law.
  3. Solve.

A free body diagram (FBD) is a problem-solving tool that we’ll use to bridge the gap between the very visual and intuitive picture, and the more abstract mathematical representation. To draw a free body diagram, represent your object as a single point, then draw arrows coming from that point that represent the different forces acting on that particular object. You also need to include coordinate axes for reference.

Example 9.3

You are designing an elevator and need to determine the tension in the cable. Your elevator will have a mass of 1200 kg (including passengers) at it’s maximum load, and it needs to be able to accelerate at 1.5 m/s2. Under these conditions, what is the tension in the cable?

The drawing should be pretty simple (it’s just a person standing in an elevator), so I’ll start with a free body diagram for the elevator. The elevator, and everything inside of it, is represented by a single point, and all the force acting on it are represented by arrows. In this case, there are only two forces: the tension in the cable, and the weight of the elevator.

A dot has two arrows extending from it. One arrow points directly up and is labeled with F superscript T. The other arrow is slightly shorter, points directly down, and is labeled with F superscript G. To the left is a short arrow pointing directly up labeled +y.
Free body diagram for the elevator

Note that the acceleration of the elevator does not appear on the free body diagram. Remember that acceleration is a description of an object’s motion—not a force—and a free body diagram only shows the forces which are acting.

Now, apply Newton’s second law and solve:

F_\textit{net,y} &= ma_y \\
F^T – F^G &= ma_y \\
F^T – mg &= ma_y \\
F^T &= ma_y + mg \\
&= m(a_y + g) \\
&= (1200\ \textrm{kg})(1.5 + 9.81)\ \textrm{m/s}^2 \\
&= 13.6\ \textrm{kN}

The cable you choose for your elevator must be able to withstand a tension of at least 13.6 kN without breaking.

Practice 9.3

Practice 9.4

Hint: you don’t need to do any math for this problem.

Practice 9.5