Questions: 3. II How close to the right edge of the 56 kg picnic table shown in Figure P8.3 can a 70 kg man stand without the table tipping over? Hint: When the table is just about to tip, what is the force of the ground on the table's left leg? FIGURE P8.3

3. II How close to the right edge of the 56 kg picnic table shown in Figure P8.3 can a 70 kg man stand without the table tipping over? Hint: When the table is just about to tip, what is the force of the ground on the table's left leg?

FIGURE P8.3

Transcript text: 3. II How close to the right edge of the 56 kg picnic table shown in Figure P8.3 can a 70 kg man stand without the table tipping over? Hint: When the table is just about to tip, what is the force of the ground on the table's left leg? FIGURE P8.3

Solution

Solution Steps

Step 1: Setting up the torque balance equation

When the table is about to tip, the normal force on the left leg is zero. The only forces acting on the table are the weight of the table, the weight of the man, and the normal force on the right leg. We can consider the pivot point to be the right leg. The net torque about this point must be zero for equilibrium. The torque due to the table's weight will act at the center of the table, 2.10 m / 2 = 1.05 m from the right edge. Let _x_ be the distance of the man from the right edge.

Step 2: Calculating the torque due to the table's weight

The torque due to the table's weight is given by:

τ_table = m_table * g * (1.05 m - 0.55 m) = 56 kg * 9.8 m/s² * 0.5 m = 274.4 Nm

Step 3: Calculating the torque due to the man's weight

The torque due to the man's weight is given by:

τ_man = m_man * g * x = 70 kg * 9.8 m/s² * x = 686x Nm

Step 4: Setting the torques equal

Since the net torque is zero, the torques must be equal and opposite:

τ_table = τ_man

274.4 Nm = 686x Nm

Step 5: Solving for x

x = 274.4 Nm / 686 N = 0.4 m