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A fun carnival ride is the wall of death, where you stand vertically (or near vertically) in a spinning cylinder and the floor drops away, but you don't fall, you're "stuck" to the wall. Here, the 80-kg person lies against the wall for which the coefficient of static friction is 4 = 0.85. Find the normal and frictional forces the wall exerts on them due to rotation about the z-axis. They have a constant speed v = 2.8 m/s. Neglect the size of the person and take 0-75°.

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A fun carnival ride is the wall of death, where you stand vertically (or near vertically) in a spinning cylinder and the floor drops away, but you don't fall, you're "stuck" to the wall. Here, the 80-kg person lies against the wall for which the coefficient of static friction is 4 = 0.85. Find the normal and frictional forces the wall exerts on them due to rotation about the z-axis. They have a constant speed v = 2.8 m/s. Neglect the size of the person and take 0-75°.

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Asked by: vishal
Created at: 2025-09-19 10:14:10

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# Wall of Death: Forces Analysis A person of mass \( m = 80\,\mathrm{kg} \) stands against the wall of a spinning cylinder. The friction between the wall and the person keeps them from sliding down when the floor drops away. ## Given Values - **Mass:** \( m = 80\,\mathrm{kg} \) - **Coefficient of static friction:** \( \mu_s = 0.85 \) - **Speed:** \( v = 2.8\,\mathrm{m/s} \) - **Angle from vertical:** \( \theta = 75^\circ \) (assume this is the angle between the normal force and vertical) - **Neglect person's size** ## Forces Acting on the Person 1. **Gravity (\( F_g \)):** Downward, \( F_g = mg \) 2. **Normal force (\( N \)):** Perpendicular to the wall (radially inward) 3. **Friction force (\( f \)):** Upward, along the wall, opposes gravity ## Free Body Diagram ``` | / | / | / <-- wall (at 75° from vertical) N |/ ------●------> (person) ^ | f (upward) ``` ## Step 1: Centripetal Force The normal force provides the centripetal acceleration: \[ N = \frac{mv^2}{r} \] But we are not given \( r \) (radius). Let's see if we can find \( N \) without it, using the vertical force balance. ## Step 2: Vertical Force Balance Set upward friction equal to downward gravity: \[ f = mg \] But friction is: \[ f = \mu_s N \] So: \[ \mu_s N = mg \implies N = \frac{mg}{\mu_s} \] **Plug in numbers:** \[ N = \frac{80 \times 9.8}{0.85} \approx \frac{784}{0.85} \approx 922\,\mathrm{N} \] ## Step 3: Frictional Force \[ f = \mu_s N = 0.85 \times 922 \approx 784\,\mathrm{N} \] ## Step 4: Check Consistency We used the minimum normal force to keep the person from sliding. If the normal force is larger (due to higher speed or smaller radius), friction would be more than enough. At the minimum speed to "stick," the frictional force exactly balances gravity. ## Step 5: Angle Consideration If the wall is at \( 75^\circ \) from vertical, the analysis above still holds, as the friction acts upward along the wall and gravity acts downward. The normal force is perpendicular to the wall, but its magnitude is still determined by the necessary friction to balance gravity. ## **Summary** ### **Normal Force** \[ \boxed{N = 922\,\mathrm{N}} \] ### **Frictional Force** \[ \boxed{f = 784\,\mathrm{N}} \] --- **Image:** ![Person in spinning cylinder with normal and friction forces](https://upload.wikimedia.org/wikipedia/commons/thumb/8/80/Wall_of_death.svg/500px-Wall_of_death.svg.png) *Alt text: Diagram of a person pressed against the wall of a spinning cylinder, with normal and friction forces indicated.* --- ### **Key Equations Used** - \( N = \frac{mg}{\mu_s} \) - \( f = \mu_s N = mg \) --- #### **Note** If the actual speed or radius gives a normal force greater than this minimum, friction would be more than enough to keep the person from falling. The above values are for the threshold where friction just balances gravity.

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