A non-uniform sign is 0.80 m long and has a weight of 18 N - AQA - A-Level Physics - Question 21 - 2018 - Paper 1

To find the extension of spring Q, we can start by analyzing the forces acting on the sign.

Since the sign is in equilibrium, the sum of the forces must equal zero. The downward force from the weight of the sign is balanced by the upward force from the combined extensions of the springs P and Q.

Let the extensions of spring P and spring Q be $x_P$ and $x_Q$, respectively. According to Hooke's law:

$F = k imes x$

Where:

• $F$ is the force (in N),
• $k$ is the spring constant (in N/m),
• $x$ is the extension (in m).

For spring P: $F_P = k_P imes x_P = 240 imes x_P$
For spring Q: $F_Q = k_Q imes x_Q = 240 imes x_Q$

The total force acting on the sign can be expressed as: $F_P + F_Q = 18 ext{ N}$
This means: $240 x_P + 240 x_Q = 18$

Thus, we have: x_P + x_Q = rac{18}{240} = 0.075 ext{ m}

Now, since the sign is in horizontal equilibrium, we can also establish a balance of moments about the center of mass of the sign (which is at the center of the 0.80 m span). Given that the distance to spring P is 0.65 m and to spring Q is 0.15 m, we can derive:

$F_P imes 0.65 = F_Q imes 0.15$
Substituting $F_P$ and $F_Q$ we get: $240 x_P imes 0.65 = 240 x_Q imes 0.15$ This simplifies to: $x_P imes 0.65 = x_Q imes 0.15$ Thus: x_P = rac{x_Q imes 0.15}{0.65} = rac{3}{13} x_Q

Now we can substitute $x_P$ back into our earlier equation: rac{3}{13} x_Q + x_Q = 0.075 Combine and solve for $x_Q$: rac{16}{13} x_Q = 0.075 Thus: x_Q = rac{0.075 imes 13}{16} = 0.061 ext{ m}

Therefore, the extension of spring Q is 0.061 m.