Two children on opposite sides of a 13 kg door (I = 1/3 mL 2 ) apply a force to the door. The first child pushes with force F at an angle of 80° relative to the door at a position r 1 from the door’s hinges. The second child pushes with a force of 15 N perpendicular to the door at a position 0.3 m from the door’s hinges. The width of the door is 0.4 m. What is the door’s angular acceleration?
R1=0.3 m
F= 20 N
To find the door's angular acceleration, we can use the principle of torque, which states that the net torque acting on an object is equal to the moment of inertia times the angular acceleration.
The net torque is the sum of the torques applied by each child. The torque applied by the first child can be calculated as:
τ1 = F * r1 * sin(θ1)
where F is the force applied by the first child, r1 is the distance from the hinge to the point where the force is applied (0.3 m), and θ1 is the angle between the force and a line drawn perpendicular to the door (80°).
The torque applied by the second child can be calculated as:
τ2 = F2 * r2
where F2 is the force applied by the second child (15 N), and r2 is the distance from the hinge to the point where the force is applied (0.3 m). Since the angle between the force and the line drawn perpendicular to the door is 90°, sin(90°) = 1, so we don't need to include it in the equation.
The total torque is given by:
τ_total = τ1 + τ2
The moment of inertia of a door rotating about its hinges can be calculated as:
I = m * L^2 / 3
where m is the mass of the door (13 kg) and L is the length of the door (0.4 m).
Finally, we can find the angular acceleration using the equation:
τ_total = I * α
where τ_total is the total torque and α is the angular acceleration we're trying to solve for.
Now, let's plug in the values:
F = 20 N
r1 = 0.3 m
θ1 = 80°
F2 = 15 N
r2 = 0.3 m
m = 13 kg
L = 0.4 m
First, let's calculate the torques:
τ1 = F * r1 * sin(θ1) = 20 * 0.3 * sin(80°) ≈ 5.76 Nm
τ2 = F2 * r2 = 15 * 0.3 ≈ 4.5 Nm
Next, let's calculate the moment of inertia:
I = m * L^2 / 3 = 13 * 0.4^2 / 3 ≈ 0.693 kgm^2
Finally, let's calculate the total torque:
τ_total = τ1 + τ2 = 5.76 + 4.5 ≈ 10.26 Nm
Now we can solve for the angular acceleration:
τ_total = I * α
10.26 = 0.693 * α
Solving for α, we get:
α = 10.26 / 0.693 ≈ 14.8 rad/s^2
Therefore, the angular acceleration of the door is approximately 14.8 rad/s^2.