You are a stickperson of mass 81.88 kg with an elastic arm as shown below. Standing on an inclined frictionless ramp of angle θ1 = 17°, you hold onto a blue rectangle to keep from sliding all the way to the bottom. You are able to find a point on the incline where you can remain at rest. At this position, your arm has stretched by 3.84 m. How much does your arm stretch when the angle of the ramp is θ2 = 64°?
I'm confused about this question. Thank you!
To address this problem, we can think in terms of the force balance at equilibrium. When the stickperson is at rest on the inclined ramp, the force due to gravity (which acts vertically downward) can be broken down into two components: one parallel to the ramp's surface and the other perpendicular to it.
Since the ramp is frictionless, the only reason the stickperson doesn't slide down is because of the tension in the elastic arm, which balances the gravitational force component parallel to the ramp.
Let's define the following:
- \( m \) = mass of the stickperson = 81.88 kg
- \( g \) = acceleration due to gravity ≈ 9.81 m/s²
- \( \theta_1 \) = initial ramp angle = 17°
- \( x_1 \) = initial stretch of the arm = 3.84 m
- \( \theta_2 \) = new ramp angle = 64°
- \( x_2 \) = new stretch of the arm (what we want to find)
- \( k \) = spring constant of the elastic arm
The tension in the arm (which equals the spring force) can be described by Hooke's Law: \( F = kx \), where \( x \) is the stretch of the arm.
At initial equilibrium, the force balance can be stated as:
\[ mg \sin(\theta_1) = kx_1 \]
The spring constant \( k \) can be determined using this equation:
\[ k = \frac{mg \sin(\theta_1)}{x_1} \]
We can plugin the known values to find \( k \):
\[ k = \frac{81.88 \cdot 9.81 \cdot \sin(17°)}{3.84} \]
\[ k = \frac{81.88 \cdot 9.81 \cdot \sin(17°)}{3.84} \approx \frac{800.4648 \cdot 0.2924}{3.84} \]
\[ k \approx \frac{234.0738}{3.84} \]
\[ k \approx 60.952 Kb \]
Now, when the ramp angle is changed to \( \theta_2 \), the new force balance at equilibrium is:
\[ mg \sin(\theta_2) = kx_2 \]
We know \( k \) and we want to find \( x_2 \), so we rearrange this to solve for \( x_2 \):
\[ x_2 = \frac{mg \sin(\theta_2)}{k} \]
Now plug in the \( k \) we found earlier and the new angle \( \theta_2 \):
\[ x_2 = \frac{81.88 \cdot 9.81 \cdot \sin(64°)}{60.952} \]
\[ x_2 = \frac{802.5348 \cdot 0.8988}{60.952} \]
\[ x_2 \approx \frac{721.5361}{60.952} \]
\[ x_2 \approx 11.839 m \]
So, when the angle of the ramp is \( \theta_2 = 64° \), the stickperson's arm would stretch approximately 11.839 meters. Keep in mind that this is a simplified physical model; real materials would have limits to their elasticity and might not behave purely according to Hooke's Law over such a large range of deformation.