A ruler stands vertically against a wall. It is given a tiny impulse at θ=0∘ such that it starts falling down under the influence of gravity. You can consider that the initial angular velocity is very small so that ω(θ=0∘)=0. The ruler has mass m= 200 g and length l= 20 cm. Use g=10 m/s2 for the gravitational acceleration, and the ruler has a uniform mass distribution. Note that there is no friction whatsoever in this problem. (See figure)
(b) What is the force exerted by the wall on the ruler when it is at an angle θ=30∘? Express your answer as the x component Fx and the y component Fy (in Newton)
Fx=
unanswered
Fy=
unanswered
(c) At what angle θ0 will the falling ruler lose contact with the wall? (0≤θ0≤90∘; in degrees) [hint: the ruler loses contact with the wall when the force exerted by the wall on the ruler vanishes.]
θ0=
To find the force exerted by the wall on the ruler when it is at an angle of θ=30∘, we can use the principles of rotational motion and equilibrium.
First, let's determine the distance of the center of mass of the ruler from the wall when it is at θ=30∘. The ruler is 20 cm long, so at θ=30∘, the vertical distance from the wall to the center of mass can be found using trigonometry:
y = l * sin(θ)
= 0.2 m * sin(30∘)
= 0.1 m
Similarly, the horizontal distance from the wall to the center of mass can be found as:
x = l * cos(θ)
= 0.2 m * cos(30∘)
= 0.173 m
Now, let's consider the forces acting on the ruler when it is at θ=30∘.
1. The weight of the ruler acts vertically downward and can be calculated as:
mg = 0.2 kg * 10 m/s^2
= 2 N
2. The force exerted by the wall can be broken down into horizontal and vertical components, Fx and Fy. Since the ruler is in equilibrium, the sum of the forces in the horizontal and vertical directions must be zero.
In the vertical direction:
Fy - mg = 0
Fy = mg
In the horizontal direction, there is no external force acting, so:
Fx = 0
Therefore, Fx and Fy can be calculated as follows:
Fx = 0 N
Fy = mg = 2 N
So the force exerted by the wall on the ruler when it is at θ=30∘ is Fx = 0 N and Fy = 2 N.
To find the angle θ0 at which the falling ruler loses contact with the wall, we need to determine the point where the force exerted by the wall becomes zero.
Since the ruler loses contact when the vertical component of the force becomes zero, we set Fy = 0 and solve for the angle θ:
Fy = mg = 0
2 N = 0.2 kg * 10 m/s^2
θ = arcsin(0) = 0∘
Therefore, the falling ruler loses contact with the wall at an angle of θ0 = 0∘.