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

a) are muscles that have bands.

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