# Physics Ninja Warrior: Ladder and Cable Problem ## Given - Mass of ladder, \( m_l = 100\,\mathrm{kg} \) - Mass of platform, \( m_p = 50\,\mathrm{kg} \) - of ladder, \( L = 2.5\,\mathrmm} \) - Platform radius, \( R = 1.\,\mathrm{m} \) - Maximum cable tension, \( T_{max} = 250\,\mathrm{N} \) - Mass of warrior, \( m_w \) (for part b, \( m_w = 72\,\mathrm{kg} \)) - Angle between cable and ladder, \( \theta \) (for part b, \( \theta = 52^\circ \)) - \( g = 9.8\,\mathrm{m/s^2} \) - \( x \) is distance from wall to warrior --- ## Part (a): Write an equation for \( x \) ### **Step 1: Draw the Force and Torque Diagram** - Hinge at wall (left end of ladder) - Ladder extends horizontally to right; length \( L \) - Cable attaches at far end (right end) at angle \( \theta \) above ladder - Warrior stands at distance \( x \) from wall (on ladder or on platform, but supported by ladder) - Platform's mass acts at its center, which is at \( L + \frac{R}{2} \) from wall ### **Step 2: Write the Torque Balance Equation About the Hinge** - Taking torques about the hinge, set counterclockwise positive. - The cable tension provides a **counterclockwise** torque: \( T \cdot L \cdot \sin\theta \) - The weights provide **clockwise** torques: - Ladder: \( m_l g \) acts at \( L/2 \) from hinge - Warrior: \( m_w g \) at \( x \) from hinge - Platform: \( m_p g \) at \( L + \frac{R}{2} \) from hinge **Torque balance:** \[ T \cdot L \cdot \sin\theta = m_l g\left(\frac{L}{2}\right) + m_w g x + m_p g \left( L + \frac{R}{2} \right) \] **Maximum value for \( T \) is \( T_{max} \):** \[ T_{max} \cdot L \cdot \sin\theta = m_l g\left(\frac{L}{2}\right) + m_w g x + m_p g \left( L + \frac{R}{2} \right) \] **Solve for \( x \):** \[ x = \frac{T_{max} L \sin\theta - m_l g \frac{L}{2} - m_p g (L + \frac{R}{2})}{m_w g} \] ### **Explanation** This equation gives the maximum distance \( x \) (from the wall) the warrior can move before the cable reaches its maximum tension. --- ## Part (b): Calculate \( x \) for Given Values Given: - \( m_w = 72\,\mathrm{kg} \) - \( \theta = 52^\circ \) ### **Step 1: Plug in All Known Values** - \( T_{max} = 250\,\mathrm{N} \) - \( L = 2.5\,\mathrm{m} \) - \( \sin\theta = \sin 52^\circ \approx .788 \) - \( m_l = 100\,\mathrm{kg} \) - \( m_p = 50\,\mathrm{kg} \) - \( R = 1.\,\mathrm{m} \) - \( g = 9.8\,\mathrm{m/s^2} \) - \( m_w = 72\,\mathrm{kg} \) Plug into the equation: \[ x = \frac{ 250 \times 2.5 \times .788 - 100 \times 9.8 \times 1.25 - 50 \times 9.8 \times (2.5 + .5) }{ 72 \times 9.8 } \] ### **Step 2: Evaluate Numerically** Calculate each part: - \( 250 \times 2.5 \times .788 = 250 \times 1.97 = 4925 \) - \( 100 \times 9.8 \times 1.25 = 100 \times 12.25 = 1225 \) - \( 50 \times 9.8 \times 3. = 50 \times 29.4 = 147 \) - Denominator: \( 72 \times 9.8 = 705.6 \) So, \[ x = \frac{4925 - 1225 - 147}{705.6} = \frac{223}{705.6} \approx 3.16\,\mathrm{m} \] ### **Explanation** This means the warrior can go up to **3.16 m** from the wall before the cable breaks. This is beyond the ladder length (\( L = 2.5\,\mathrm{m} \)), which means the cable will not break even if you walk to the end of the ladder and onto the platform a little. --- ## **Final Answers** ### **(a) Equation for \( x \):** \[ \boxed{ x = \frac{T_{max} L \sin\theta - m_l g \frac{L}{2} - m_p g (L + \frac{R}{2})}{m_w g} } \] ### **(b) Numerical Value for \( x \):** \[ \boxed{ x \approx 3.16\,\mathrm{m} } \] This is the maximum distance from the wall the warrior can go before the cable breaks.