Question

additional practice

1. ashok, whose mass is 43 kg, sits 1.8 m from a pivot at the center of a seesaw. steve, whose mass is 52 kg, wants to seesaw with ashok. how far from the center of the seesaw should steve sit?
2. a bicycle - chain wheel has a radius of 7.70 cm. if the chain exerts a 35.0 - n force on the wheel in the clockwise direction, what torque is needed to keep the wheel from turning?

Explanation:

Problem 16: Seesaw Equilibrium (Torque Balance)

Step 1: Recall Torque Balance Condition

For the seesaw to be in equilibrium, the clockwise torque must equal the counterclockwise torque. Torque \(\tau = rF\sin\theta\), and for a seesaw, \(\theta = 90^\circ\) (force is perpendicular to the lever arm), so \(\sin\theta = 1\). The force here is the weight \(F = mg\). Let \(r_A\) be Ashok's distance from the pivot, \(m_A\) his mass, \(r_S\) Steve's distance, and \(m_S\) his mass. Then \(m_A g r_A = m_S g r_S\). The \(g\) cancels out.

Step 2: Plug in Values

We know \(m_A = 43\ \text{kg}\), \(r_A = 1.8\ \text{m}\), \(m_S = 52\ \text{kg}\). Solve for \(r_S\):
\(r_S = \frac{m_A r_A}{m_S} = \frac{43\ \text{kg} \times 1.8\ \text{m}}{52\ \text{kg}}\)

Step 3: Calculate

\(r_S = \frac{77.4}{52} \approx 1.49\ \text{m}\)

Step 1: Recall Torque Formula

Torque \(\tau = rF\sin\theta\). The force from the chain is tangential to the wheel, so \(\theta = 90^\circ\) and \(\sin\theta = 1\). The radius \(r\) is \(7.70\ \text{cm} = 0.0770\ \text{m}\), force \(F = 35.0\ \text{N}\).

Step 2: Plug in Values

\(\tau = rF = 0.0770\ \text{m} \times 35.0\ \text{N}\)

Step 3: Calculate

\(\tau = 2.695\ \text{N·m} \approx 2.70\ \text{N·m}\) (to keep the wheel from turning, the opposing torque must equal this, so magnitude is \(2.70\ \text{N·m}\)).

Answer:

Steve should sit approximately \(1.49\ \text{m}\) from the center.

Problem 17: Torque on Bicycle Wheel