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?

γ = 1.42 * F1 * r1 - 6.52 Nm

Explanation:

The first kid will apply a torque of:

T = F * d * sin(a)

T1 = F1 * r1 * sin(80)

T1 = 0.98 * F1 * r1

The second kid applies a torque of

T2 = -15 * 0.3 * sin(90) = -4.5 Nm (negative because it is in the opposite direction as the other kid)

The angular momentum of the door is:

I = 1/3 * m * L^2

I = 1/3 * 13 * 0.4^2 = 0.69 kg*m^2

The total torque applied is:

T = T1 + T2

T = 0.98 * F1 * r1 - 4.5 Nm

The angular acceleration will be:

γ = T / I

γ = (0.98 * F1 * r1 - 4.5) / 0.69

γ = 1.42 * F1 * r1 - 6.52 Nm

i think it’s a. this is because if they measured the mass of something with a mass of 150g, then their final answer would have to be approximately 150g. though all of their answers are almost 150, rosa and tran have the closest answers (or the exact answer) to the accurate one.