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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
Question 21
A non-uniform sign is 0.80 m long and has a weight of 18 N. It is suspended from two vertical springs P and Q. The springs obey Hooke's law and the spring constant ... show full transcript
Worked Solution & Example Answer: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
Step 1
What is the extension of spring Q?
Answer
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 and , respectively. According to Hooke's law:
Where:
- is the force (in N),
- is the spring constant (in N/m),
- is the extension (in m).
For spring P:
For spring Q:
The total force acting on the sign can be expressed as:
This means:
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:
Substituting and we get:
This simplifies to:
Thus:
x_P = rac{x_Q imes 0.15}{0.65} = rac{3}{13} x_Q
Now we can substitute back into our earlier equation: rac{3}{13} x_Q + x_Q = 0.075 Combine and solve for : 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.