Q11.E - University Physics with Modern Physics

Question

A diving board 3.00 m long is supported at a point 1.00 m from the end, and a diver weighing 500 N stands at the free end (Fig. E 11.11). The diving board is of uniform cross section and weighs 280 N. Find :

(a) The force at the support point and 

(b) The force at the left-hand end.

 

Step-by-Step Solution

Answer

(a) Thus, the force at the support point is 1920 N.

(b) Thus, the force at the left-hand end of the board is 1140 N.

Step 1: Equilibrium

The condition for translational equilibrium can be expressed as: $\mathbf{\sum }{\mathit{F}}_{\mathbf{e}\mathbf{x}\mathbf{t}}\mathbf{=}\mathbf{0}$ .

And that for rotational equilibrium can be expressed as: $\mathbf{\sum }{\mathit{\tau }}_{\mathbf{e}\mathbf{x}\mathbf{t}}\mathbf{=}\mathbf{0}$ . The sum of all the forces acting on the body will be zero.

Step 2: a. Find the Force

Given that the diver standing at one end of the diving board weighs 500 N. The weight and length of the board, respectively, are 280 N and 3.00 m.

Let the whole given setup be illustrated as a free body diagram for forces on the board, as shown in the figure below:

Now, for the force $F_1$, considering the criteria for rotational equilibrium, we have

$$\begin{gathered} \sum \tau =0 \\ {F}_1\left(1.00\right)=\left(1.5\right)\left(280\right)+\left(3.00\right)\left(500\right) \\ F_1=1920 \text{N} \end{gathered}$$

Thus, the force at the support point is 1920 N.

Step 3: b. Find the Force

Similarly, for the force $F_2$, considering the criteria for rotational equilibrium at the other end, we have:

$$\begin{gathered} \sum \tau =0 \\ {F}_2\left(1.00\right)=\left(0.50\right)\left(280\right)+\left(2.00\right)\left(500\right) \\ F_2=1140 \text{N} \end{gathered}$$

Thus, the force at the left-hand end of the board is 1140 N..