A force of $10$ N is applied at the farthest edge from the hinge of the door of width $50$ cm such that the line of action of force makes angle of $90^{o}$ with the normal to the surface of the door. Determine the moment of the force of the door.

5

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500

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zero

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50

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Solution

The correct option is C zero
Since the force makes $90^{o}$ angle to the normal to the surface of door, thus angle between the force applied and the position vector is $0^{o}$ .
Net torque $t a u = r F sin θ$
Since, $θ = 0^{o}$ $⟹ τ = 0$

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A force of 10 N is applied at the farthest edge from the hinge of the door of width 50 cm such that the line of action of force makes angle of 90o with the normal to the surface of the door. Determine the moment of the force of the door.

The correct option is C zeroSince the force makes 90o angle to the normal to the surface of door, thus angle between the force applied and the position vector is 0o.Net torque tau=rFsinθSince, θ=0o ⟹ τ=0

A force of $10$ N is applied at the farthest edge from the hinge of the door of width $50$ cm such that the line of action of force makes angle of $90^{o}$ with the normal to the surface of the door. Determine the moment of the force of the door.

The correct option is C zero
Since the force makes $90^{o}$ angle to the normal to the surface of door, thus angle between the force applied and the position vector is $0^{o}$ .
Net torque $t a u = r F sin θ$
Since, $θ = 0^{o}$ $⟹ τ = 0$