Questions: Refer to the procedure, Part 1, 1st arrangement. Assume xcm=50.0 cm, 150.0 g is suspended from a hanger clamp at the position xcc=15.0 cm, and a hanger clamp is at position xc=75.0 cm. If each hanger clamp has a mass m=16.5 g, what mass must be added to xc in order to attain stable equilibrium? Sketch a diagram of the situation (refer to Fig. 8.3). ( 30 pts ) Consider Part 2 of the procedure. Determine the additional mass required for stable equilibrium. Meter stick: xcm=50.0 cm, m=150.0 g. Hanger clamp: xcc=0.0 cm, m=16.5 g .(30 pts)

Solution Steps

First, draw a free body diagram of the meter stick. The forces acting on the meter stick are: the weight of the meter stick (W_m) at x_cm = 50 cm, the weight of the mass (W) at x_c = 75 cm, the weight of the clamp holding the mass (W_clamp) also at x_c = 75cm, the weight of the other clamp (W_clamp) at x_cc = 15 cm, and the unknown mass (W_unk) at the same location, x_cc = 15cm. We need to balance the torques around the fulcrum (at x = x_cm = 50 cm) for equilibrium.

The torque is the force times the perpendicular distance to the pivot point. Clockwise torques will be considered positive, and counterclockwise torques will be considered negative. Our equation for rotational equilibrium is:

∑τ = (W + W_clamp)*(x_c - x_cm) + W_clamp * (x_cc - x_cm) – W_unk * (x_cm – x_cc) = 0

Substituting the weights with masses (W = mg) gives:

(m_g + m_clamp_g)_(x_c - x_cm) + m_clamp_g * (x_cc - x_cm) – m_unk * g* (x_cm – x_cc) = 0

We can cancel _g_ from the equation:

(m + m_clamp)*(x_c - x_cm) + m_clamp * (x_cc - x_cm) – m_unk * (x_cm – x_cc) = 0

Plug in the known values: m = 150 g, m_clamp = 16.5 g, x_c = 75 cm, x_cc = 15 cm, x_cm = 50 cm

(150 g + 16.5 g)*(75 cm - 50 cm) + 16.5 g * (15 cm - 50 cm) – m_unk * (50 cm – 15 cm) = 0

(166.5 g)*(25 cm) + 16.5 g * (-35 cm) - m_unk * (35 cm) = 0

4162.5 g_cm – 577.5 g_cm = m_unk * 35 cm

m_unk = (3585 g*cm) / (35 cm)

m_unk = 102.4 g

Final Answer: