Rotational Equilibrium and Newton's First Law Practice Test

A balanced mobile uses a light horizontal bar pivoted at its center, staying at rest. A 2.0 N weight hangs 0.40 m left of the pivot, and a 1.0 N weight hangs 0.10 m right of the pivot. A third weight of 1.5 N is moved along the right side to reestablish equilibrium. Each weight exerts a downward force vector, and each produces torque equal to $rF$ about the pivot. The system is in rotational equilibrium, so the sum of clockwise and counterclockwise torques is zero, consistent with Newton’s First Law in rotational form. Considering the forces acting on the object, where should the 1.5 N weight be placed (distance from the pivot on the right) to maintain equilibrium?

A seesaw pivots at its center and stays level while two students sit on opposite sides. A $300,\text{N}$ student sits $1.2,\text{m}$ to the left of the pivot, and a second student sits to the right with an unknown weight $W$ at a distance $0.80,\text{m}$. The forces act vertically downward at their seats, creating torques with lever arms measured from the pivot. The support force at the pivot produces no torque about the pivot. The seesaw is motionless, so Newton’s First Law in rotational form requires $\sum \tau_{\text{pivot}}=0$. Considering the forces acting on the seesaw, calculate the force needed at a specific point to maintain equilibrium.