Question

A uniform plank rests on a level surface as shown below. The plank has a mass of 30 kg and is 6.0 m long. How much mass can be placed at its right end before it tips? (Hint: When the board is about to tip over, it makes contact with the surface only along the edge that becomes a momentary axis of rotation.)

Answer 1

To prevent the uniform plank from tipping, the maximum additional mass that can be placed at the right end is 15 kg, which is found by equating the torque from the plank's weight to the torque from the added mass about the momentary rotation at the left edge.

Step-by-step explanation:

The plank mentioned is in a state of static equilibrium when it is just about to tip over. Since the plank is uniform, its center of gravity is at its center, which is 3.0 m from either end. The weight of the plank (Wplank) acts downward at this point, and the weight of the added mass (Wmass) acts at the right end of the plank.

To find how much mass can be placed at the right end before the plank tips, we must ensure that the torque due to the weight of the plank about the pivot (the left edge of the plank) is equal to the torque due to the added mass. The torque (τ) is the product of force (F) and the perpendicular distance from the pivot to the line of action of the force (r), that is, τ = r × F.

The torque due to the plank's weight about the pivot is τplank = (3.0 m) × (30 kg × 9.8 m/s2). The momentary axis of rotation is at the edge where the plank makes contact with the surface, so the torque due to the added mass at the edge of the plank before tipping would be τmass = (6.0 m) × (mass × 9.8 m/s2). Setting τplank equal to τmass allows us to solve for the mass. Therefore, τplank = τmass gives us (3.0 m) × (30 kg × 9.8 m/s2) = (6.0 m) × (mass × 9.8 m/s2), and when simplified, the mass = (3.0 m / 6.0 m) × 30 kg, which equals 15 kg.

Answer 2

To determine how much mass can be placed at the right end of the plank before it tips, we need to consider the concept of torque. The maximum mass that can be placed at the right end before tipping is zero. This means that no additional mass can be added without tipping the plank.

Step-by-step explanation:

To determine how much mass can be placed at the right end of the plank before it tips, we need to consider the concept of torque. Torque is the product of force and the perpendicular distance from the point of rotation. In this case, the point of rotation is the left end of the plank.

For the plank to be in static equilibrium, the total clockwise torque acting on the plank must be equal to the total counterclockwise torque. The clockwise torque is produced by the weight of the plank (30 kg * 9.8 m/s² * 6.0 m) and the mass at the right end (m kg * 9.8 m/s² * x m). The counterclockwise torque is produced by the weight of the plank (30 kg * 9.8 m/s² * 6.0 m).

To find the maximum mass that can be placed at the right end before tipping, we set the clockwise and counterclockwise torques equal to each other and solve for the mass:

(30 kg * 9.8 m/s² * 6.0 m) + (m kg * 9.8 m/s² * x m) = 30 kg * 9.8 m/s² * 6.0 m

Simplifying the equation, we get:

m kg * x m = 0

Since the left side of the equation is zero, the maximum mass that can be placed at the right end before tipping is zero. This means that no additional mass can be added without tipping the plank.