Question

a mass m is at the end of a horizontal plank of mass m and length l. the plank is supported at two points, at 0 and at 2/3l and friction is sufficient so that it would not slip on those points. a) determine the maximum mass m that can be at rest, in terms of m. b) for a mass m twice that value, determine the angular acceleration of the plank, as it starts rotating, in terms of g and l.

Answer 1

The maximum mass m that can be at rest on the plank without it tipping is m = 2/3l. The angular acceleration of the plank when a mass m twice the maximum value is placed at the end is α = (3/4)g/l.

Step-by-step explanation:

In order to determine the maximum mass m that can be at rest on the plank without it tipping, we need to consider the torque acting on the plank. The torque due to the weight of the mass m is given by the product of the mass, the acceleration due to gravity, and the distance from the pivot point. Taking moments about the left support, we have:

m × g × (2/3l) = m × g × l

Simplifying, we find that the maximum mass m that can be at rest on the plank is m = 2/3l.

To determine the angular acceleration of the plank when a mass m twice the maximum value is placed at the end, we need to consider the net torque acting on the system. The torque due to the weight of the mass m is given by m × g × (2/3l). The torque due to the weight of the additional mass m is given by m × g × l. The total torque is the sum of these torques. The moment of inertia of the plank about the left support is given by (1/3)m × l^2. Using Newton's second law for rotation, we have:

(3/4)m × g × l = I × α

Simplifying, we find that the angular acceleration α of the plank is α = (3/4)g/l.