WPE BYU 31

Solution 31

Solution to Question 31

Mid Hinge m Tension T dy 2dy

1. Geometric Analysis

The pantograph consists of two identical rhombi stacked vertically. Let the vertical height of one rhombus be $h$.

  • Height of middle hinge: $y_{mid} = h$
  • Height of lowest hinge (load): $y_{load} = 2h$

For any small virtual displacement, the relationship is linear:

$$\delta y_{load} = 2 \delta y_{mid}$$

2. Principle of Virtual Work

We apply a virtual displacement downwards. The total work done by all active forces must be zero.

  • Work by Gravity ($W_g$): Force $mg$ acts on the load moving by $\delta y_{load}$. $$W_g = mg \cdot \delta y_{load}$$
  • Work by Cord Tension ($W_T$): The cord connects the middle and lowest hinges. The tension $T$ opposes the separation of these points. The separation increases by $\delta y_{load} – \delta y_{mid}$. $$W_T = -T (\delta y_{load} – \delta y_{mid})$$

3. Solve for Tension

Sum of work is zero:

$$mg \delta y_{load} – T (\delta y_{load} – \delta y_{mid}) = 0$$

Substitute $\delta y_{load} = 2 \delta y_{mid}$:

$$mg (2 \delta y_{mid}) – T (2 \delta y_{mid} – \delta y_{mid}) = 0$$ $$2 mg \delta y_{mid} – T \delta y_{mid} = 0$$ $$T = 2mg$$