Suppose a door is 1 meter wide. Calculate the difference in the amount torque exerted on the door when someone pushes with a constant force of 100N at the handle side of the door compared to the middle of the door. Assume the force is perpendicular to the surface of the door.

Δτ = 50 N.m

Explanation:

The torque applied on an object is given by the product of the force applied on it and the perpendicular distance between the force and the axis of rotation of the object. That is:

τ = F r

where,

τ = Torque applied on the object

F = Force applied on it

r = distance from axis of rotation

FOR HANDLE SIDE OF DOOR:

τ₁ = F r₁

where,

τ₁ = Torque applied on the object = ?

F = Force applied on it = 100 N

r₁ = distance from axis of rotation = 1 m

Therefore,

τ₁ = (100 N)(1 m)

τ₁ = 100 N.m

FOR MIDDLE OF DOOR:

τ₂ = F r₂

where,

τ₂ = Torque applied on the object = ?

F = Force applied on it = 100 N

r₂ = distance from axis of rotation = 1 m/2 = 0.5 m

Therefore,

τ₂ = (100 N)(0.5 m)

τ₂ = 50 N.m

Now, the difference between the amount of torque in both cases is:

Δτ = τ₁ - τ₂

Δτ = 100 N.m - 50 N.m

Δτ = 50 N.m

pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y

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