The torque is given by $$\tau = rF\sin\phi$$, where $$r$$ is the distance from the pivot to the point of force application, $$F$$ is the force, and $$\phi$$ is the angle between the position vector and the force vector. Here, the pivot is the base of the ladder, $$r=L$$, and $$F = N_w$$. The angle between the ladder (position vector) and the horizontal force $$N_w$$ is $$(180^\circ - \theta)$$. The lever arm of the force $$N_w$$ about the base is $$L\sin\theta$$. Thus, the torque is $$N_w (L\sin\theta)$$.

For the bar to be in static equilibrium under its own weight, the pivot must provide a normal force that exactly opposes the gravitational force, and there must be zero net torque. To achieve zero net torque from gravity, the pivot must be placed directly under the center of mass. This way, the lever arm for the gravitational force is zero, resulting in zero torque.

To find $$N_1$$, we sum the torques about Support 2 ($$x=L$$) and set them to zero. The clockwise torques are from the plank's weight $$Mg$$ at $$x=L/2$$ and the block's weight $$mg$$ at $$x=L/3$$. The counter-clockwise torque is from $$N_1$$ at $$x=0$$. The lever arms relative to $$x=L$$ are $$L$$, $$L/2$$, and $$2L/3$$ respectively. So, $$N_1(L) - Mg(L/2) - mg(2L/3) = 0$$. Solving for $$N_1$$ gives $$N_1 = Mg/2 + 2mg/3$$.

According to Newton's first law for rotation, an object will maintain a constant angular velocity if and only if the net external torque acting on it is zero. The engine provides a forward torque, which is exactly balanced by the resistive torque from air drag, resulting in a net torque of zero and thus a constant angular velocity.

For rotational equilibrium, the net torque about the center of mass must be zero. The tension $$T$$ creates a torque $$Tr$$ (let's say clockwise). The static friction force $$f_s$$ acts at the point of contact with the table (radius $$R$$) and must create an equal and opposite (counter-clockwise) torque $$f_s R$$. Therefore, $$Tr = f_s R$$, which gives $$f_s = T(r/R)$$.

According to Newton's first law for rotation, if an object rotates with a constant angular velocity, its angular acceleration is zero. This implies that the net torque acting on it must be zero. The torque applied by the person must be exactly balanced by an opposing frictional torque from the hinges for the net torque to be zero.

This question tests understanding of rotational equilibrium and Newton's First Law in rotational form in AP Physics C: Mechanics. Rotational equilibrium occurs when the net torque about the pivot is zero, meaning clockwise and counterclockwise torques balance perfectly. In this scenario, the 2.0 N weight creates counterclockwise torque (2.0 × 0.40 = 0.80 N·m), while the 1.0 N weight creates clockwise torque (1.0 × 0.10 = 0.10 N·m), leaving a net counterclockwise torque of 0.70 N·m that must be balanced by the 1.5 N weight. Choice A is correct because setting up the balance equation: 0.80 = 0.10 + 1.5 × d, where d is the distance right of the pivot, gives 1.5d = 0.70, so d = 0.70/1.5 = 0.467 m ≈ 0.47 m. Choice C (0.07 m) might result from arithmetic errors, while choice D incorrectly places the weight on the left side. To help students: Create a torque inventory table listing each force, its lever arm, and resulting torque with proper signs, always verify that the sum of all torques equals zero in the final configuration, and use dimensional analysis to ensure the answer has correct units.

This question tests understanding of rotational equilibrium and Newton's First Law in rotational form in AP Physics C: Mechanics. Rotational equilibrium occurs when the sum of torques acting on an object is zero, meaning the object is not accelerating rotationally. Newton's First Law in rotational form states that an object at rest will remain so unless acted upon by a net external torque. In this scenario, we have a seesaw with two students sitting on opposite sides of a central pivot, where we need to find the unknown weight to maintain balance. Choice B is correct because applying torque balance about the pivot: clockwise torque = counterclockwise torque, so 300 N × 1.2 m = W × 0.80 m, giving W = 360/0.80 = 450 N. Choice A is incorrect because it reverses the lever arms in the calculation, using 300(0.80) = W(1.2) instead of the correct relationship. To help students: Draw a clear diagram showing the pivot, forces, and lever arms. Emphasize that torque = force × perpendicular distance from pivot, and practice setting up the torque balance equation systematically with proper signs for clockwise vs counterclockwise torques.

This question tests understanding of rotational equilibrium and Newton's First Law in rotational form in AP Physics C: Mechanics. Rotational equilibrium requires that both the net torque and net force on an object equal zero, ensuring no linear or angular acceleration. In this scenario, the seesaw experiences torques from two people: 500 N × 1.0 m = 500 N·m counterclockwise and 250 N × 2.0 m = 500 N·m clockwise, which balance perfectly. Choice A is correct because for complete static equilibrium, both conditions must be satisfied: Στ = 0 (no angular acceleration) and ΣF = 0 (no linear acceleration). Choice B is incorrect because torque balance is essential for rotational equilibrium, choice C incorrectly applies the rotational dynamics equation to a static situation, and choice D contradicts the equilibrium condition. To help students: Emphasize that static equilibrium requires two separate conditions to be met simultaneously, practice identifying all forces and their points of application, and reinforce that torque depends on both force magnitude and lever arm distance.

This question tests understanding of rotational equilibrium and Newton's First Law in rotational form in AP Physics C: Mechanics. Rotational equilibrium requires the net torque about any point to be zero, with torque depending on both the force magnitude and its perpendicular distance from the pivot. In this scenario, the cable creates counterclockwise torque through its vertical component T sin(θ), which must balance the clockwise torques from the beam and sign weights. Choice A is correct because increasing θ increases sin(θ), thereby increasing the vertical component of the tension and its torque without changing the tension magnitude T. Choice B is incorrect because cos(θ) relates to the horizontal component which doesn't contribute to torque about a horizontal axis, choice C would increase clockwise torque making balance harder, and choice D incorrectly references moment of inertia which doesn't affect static torque. To help students: Use vector decomposition to identify which force components create torque, remember that torque depends on the perpendicular distance between the force line and pivot, and practice varying one parameter while holding others constant to see the effect.