A horizontal metal peg is used to support a component on a vertical structural wall.
A specialized two-pronged wrench is used to tighten the peg. The width of the wrench from end to end is 8.0×10−3 m8.0 \times 10^{-3}\text{ m}8.0×10−3 m. The prongs exert equal and opposite forces on the peg. The magnitude of each force is 280 N280\text{ N}280 N, as shown in Fig. 22.1.
Calculate the magnitude of the torque of the couple produced by these forces.
torque=\text{torque} = torque= ............................ N m\text{N m}N m
The component of mass mmm is then hung on the peg at a point BBB, as shown in Fig. 22.2.
The inside section of the support structure exerts a maximum downward force of 45 N45\text{ N}45 N on the embedded section of the peg at a distance of 4.0×10−2 m4.0 \times 10^{-2}\text{ m}4.0×10−2 m from the outer edge of the wall. The hanging component exerts a downward force FFF on the peg at a distance of 6.0×10−3 m6.0 \times 10^{-3}\text{ m}6.0×10−3 m from the outer edge of the wall. There is an upward reaction force RRR acting on the peg at the outer edge of the support structure. The mass of the peg itself is negligible.
Use the principle of moments to calculate the maximum mass mmm of the hanging component. Take g=9.81 m s−2g = 9.81\text{ m s}^{-2}g=9.81 m s−2.
mass=\text{mass} = mass= ............................ kg\text{kg}kg
