Question

Tamar wants to cut down a large, dead poplar tree with her chain saw, but she does not want it to fall onto the nearby gazebo. Yoojin comes to help with a long rope. Yoojin, a physicist, suggests they tie the rope taut from the poplar to the oak tree and then pull sideways on the rope as shown in the figure. If the rope is 40.0 m long and Yoojin pulls sideways at the midpoint of the rope with a force of \(360.0 \mathrm{N},\) causing a \(2.00-\mathrm{m}\) sideways displacement of the rope at its midpoint, what force will the rope exert on the poplar tree? Compare this with pulling the rope directly away from the poplar with a force of \(360.0 \mathrm{N}\) and explain why the values are different. [Hint: Until the poplar is cut through enough to start falling, the rope is in equilibrium.] (IMAGE CAN'T COPY)

Short answer

Answer: The force exerted on the poplar tree when Yoojin pulls sideways at the midpoint of the rope is approximately 1792 N. This force is significantly greater than the force exerted when pulling the rope directly away from the poplar tree, which would be 360 N.

Step-by-step solution

Analyzing the problem

We have a rope 40.0 m long, and Yoojin pulls sideways at the midpoint of the rope with a force of \(360.0 \mathrm{N},\) causing a \(2.00-\mathrm{m}\) sideways displacement of the rope. We need to find the force exerted on the poplar tree. With the hint provided, we know that the rope is in equilibrium. Since forces are acting in two dimensions (horizontal and vertical), we will apply Newton's second law for both dimensions.

Calculating the angle

To analyze the tension in the rope, we must first find the angle \(\theta\) between the horizontal and the rope. When Yoojin pulls the rope sideways causing \(2.00-\mathrm{m}\) sideways displacement, we can use the following equation to find the angle \(\theta\): tan \(\theta = \frac{h}{20.0}\) tan \(\theta = \frac{2.00}{20.0}\) tan \(\theta\) = 0.1 We can then use the inverse tangent to find the angle \(\theta\): \(\theta = tan^{-1}(0.1)\) \(\theta = 5.71^{\circ}\)

Breaking the tension into components

Now that we have the angle, we can break the tension force into two components: horizontal and vertical. \(T_x = T\cos\theta\) \(T_y = T\sin\theta\) However, since we have symmetry in this problem (Yoojin pulls the rope at the midpoint), we can write that there are two equal and opposite vertical forces: \(2T_y = 360.0 \mathrm{N}\) Then, we can use the value of \(\sin\theta\): \(\sin\theta = \sin(5.71^{\circ}) = 0.0997\)

Finding the tension T

Next, we substitute \(T_y\) expression into the equilibrium condition: \(2(T\sin\theta) = 360.0\) \(T\sin\theta = \frac{360.0}{2} = 180.0\) Now, we can find the tension T: \(T = \frac{180.0}{0.0997} \approx 1805 \mathrm{N}\)

Calculating the horizontal force on the poplar tree

Now that we have the tension T, we can calculate the horizontal force on the poplar tree, which is \(T_x\): \(T_x = T\cos\theta\) \(T_x = 1805 \times \cos(5.71^{\circ})\) Calculating \(T_x\), we get: \(T_x \approx 1792 \mathrm{N}\) So the force exerted on the poplar tree is approximately \(1792 \mathrm{N}\).

Comparing with pulling the rope directly away

If Yoojin were to pull the rope directly away from the poplar tree with a force of \(360.0 \mathrm{N}\), the force exerted on the poplar tree would be the same - \(360.0 \mathrm{N}\). The reason for the difference in values is that in the case when Yoojin pulls the rope sideways at the midpoint, the tension force in the rope has a horizontal component that acts directly on the poplar tree, while the vertical components cancel out. Thus, this method helps to exert a greater force on the poplar tree with the same amount of applied force.