A plank, of length L = 3.7 m and mass M = 19.6 kg, rests on the ground and on a frictionless roller at the top of a wall of height h = 1.70 m (see Figure). The center of gravity of the plank is at its center. The plank remains in equilibrium for any value of theta => 78o but slips if theta < 78o. Calculate the magnitude of the force exerted by the roller on the plank when theta = 78o.
calculating moments about the bottom contact point:
force on wall*L/2*sin78=19.6*L/2*cos78
force on wall= 19.6 ctan78
To calculate the magnitude of the force exerted by the roller on the plank when theta = 78°, we can use the principle of torque equilibrium.
When the plank is in equilibrium, the net torque about any point is zero. We can choose any point as the pivot, but it is often convenient to choose a point where one or more forces pass through to eliminate torque contributions from those forces.
In this case, let's choose the point where the plank makes contact with the roller as the pivot. Since the roller is frictionless, the force of gravity acting on the plank's center of gravity passes through this pivot point and does not contribute to the torque about the pivot.
The torque due to the force exerted by the roller is 0 because the force passes through the pivot point.
Now, let's consider the torque due to the force of gravity acting on the plank. The force of gravity can be broken down into its components: one parallel to the plank and one perpendicular to it. The component parallel to the plank does not contribute to the torque as it passes through the pivot point. However, the perpendicular component of the force of gravity is responsible for the torque.
The perpendicular component of the force of gravity can be calculated using the equation: F_perpendicular = m * g * sin(theta), where m is the mass of the plank, g is the acceleration due to gravity, and theta is the angle between the plank and the horizontal.
Since the plank is in equilibrium, the torque due to the perpendicular component of the force of gravity must be balanced by the torque due to the force exerted by the roller.
The torque due to the perpendicular component of the force of gravity is given by: Torque_gravity = F_perpendicular * d, where d is the perpendicular distance between the pivot point and the line of action of the force.
In this case, the perpendicular distance d is equal to the height of the wall, h.
Setting the torque due to the perpendicular component of the force of gravity equal to zero, we can solve for the unknown force exerted by the roller.
Torque_gravity = Torque_roller
(F_perpendicular * d) = (force_roller * d)
m * g * sin(theta) * h = force_roller * h
Since sin(78°) = 0.978, we can substitute this value in the equation:
m * g * sin(78°) * h = force_roller * h
Plugging in the known values:
(19.6 kg) * (9.8 m/s^2) * (0.978) * (1.70 m) = force_roller * (1.70 m)
Solving for force_roller:
force_roller = (19.6 kg) * (9.8 m/s^2) * (0.978)
force_roller ≈ 183.38 N
Therefore, the magnitude of the force exerted by the roller on the plank when theta = 78° is approximately 183.38 Newtons.