A homeowner is trying to move a stubborn rock from his yard. By using a a metal rod as a lever arm and a fulcrum (or pivot point) the homeowner will have a better chance of moving the rock. The homeowner places the fulcrum a distance d=0.288 m from the rock, which has a mass of 345 kg, and fits one end of the rod under the rock’s center of weight. d- If the homeowner can apply a maximum force of 711 N at the other end of the rod, what is the minimum total length L of the rod required to move the rock? Assume that the rod is massless and nearly horizontal so that the weight of the rock and homeowner’s force are both essentially vertical.
The Correct Answer and Explanation is :
In this problem, we need to determine the minimum length of a rod that allows a homeowner to lift a rock using a lever system. The force applied at one end of the rod and the weight of the rock create torques about the fulcrum. For the rock to be lifted, the torque generated by the applied force must be at least equal to the torque generated by the weight of the rock.
Key Concepts:
- Torque (τ) is calculated using the formula:
[
\tau = F \cdot d
]
where (F) is the force applied, and (d) is the perpendicular distance from the pivot (fulcrum) to the line of action of the force. - Equilibrium condition: The torque applied at one end of the rod must balance the torque due to the weight of the rock at the other end. This can be written as:
[
F_{\text{applied}} \cdot (L – d) = W_{\text{rock}} \cdot d
]
where:
- (F_{\text{applied}}) is the maximum force the homeowner can apply (711 N).
- (W_{\text{rock}}) is the weight of the rock, which is (W = m \cdot g = 345 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 3384.45 \, \text{N}).
- (d) is the distance from the fulcrum to the center of the rock’s weight (0.288 m).
- (L) is the total length of the rod, which is what we need to solve for.
Solution:
First, plug in the known values into the equilibrium condition:
[
711 \cdot (L – 0.288) = 3384.45 \cdot 0.288
]
Solve for (L):
[
711 \cdot (L – 0.288) = 974.77
]
[
L – 0.288 = \frac{974.77}{711} \approx 1.371
]
[
L = 1.371 + 0.288 \approx 1.659 \, \text{m}
]
Conclusion:
The minimum total length of the rod required to move the rock is approximately 1.659 meters. This length ensures that the torque generated by the applied force is sufficient to lift the rock.