Answer
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Hint We know the definition of moment of force. We also know the axis of rotation. The product of the magnitude of the applied force and the perpendicular distance of the line of action of the force from the axis of rotation gives the magnitude of the moment of force. So, we can write that in equilibrium according to the principle of moments, the sum of anticlockwise moments equals the sum of clockwise moments.
Complete step by step solution
The force applied on a body and the perpendicular distance of the line of action of the force from the axis of rotation together, set up a rotational tendency of the body, and is called moment of the force about the axis of rotation.
The straight line through all fixed points of a rotating rigid body around which all other points of the body move in circles is called the axis of rotation.
Product of the magnitude of the applied force (F) and the perpendicular distance (d) of the line of action of the force from the axis of rotation gives the magnitude of the moment of force $(G) = Fd$
A number of forces acting on the same plane are called coplanar forces.
Let here all the magnitude of the forces be same = F
Moment of force ${F_1} = 0$
For ${F_2}$ the rotation of body AB is anticlockwise. As per convention, for anticlockwise rotation the moment of force is taken as positive.
For ${F_3}$ the rotation is clockwise here the moment of the force ${F_3}$ is taken as negative.
Hence the moments of coplanar forces can be expressed as ordinary positive or negative algebraical quantities.
In equilibrium according to the principle of moments, the sum of anticlockwise moments equals the sum of clockwise moments.
Hence the algebraic sum of moments of all the forces acting on the body about the axis of rotation is zero if the body in equilibrium (Option- “B”)
Note Given that the body is in equilibrium. If the sum is zero then the body will not rotate. So, option “B” is correct. If the sum is positive the body turns anticlockwise so option “C” is not correct. If negative then it turns clockwise so option “D” is also not correct.
Complete step by step solution
The force applied on a body and the perpendicular distance of the line of action of the force from the axis of rotation together, set up a rotational tendency of the body, and is called moment of the force about the axis of rotation.
The straight line through all fixed points of a rotating rigid body around which all other points of the body move in circles is called the axis of rotation.
Product of the magnitude of the applied force (F) and the perpendicular distance (d) of the line of action of the force from the axis of rotation gives the magnitude of the moment of force $(G) = Fd$
A number of forces acting on the same plane are called coplanar forces.
Let here all the magnitude of the forces be same = F
Moment of force ${F_1} = 0$
For ${F_2}$ the rotation of body AB is anticlockwise. As per convention, for anticlockwise rotation the moment of force is taken as positive.
For ${F_3}$ the rotation is clockwise here the moment of the force ${F_3}$ is taken as negative.
Hence the moments of coplanar forces can be expressed as ordinary positive or negative algebraical quantities.
In equilibrium according to the principle of moments, the sum of anticlockwise moments equals the sum of clockwise moments.
Hence the algebraic sum of moments of all the forces acting on the body about the axis of rotation is zero if the body in equilibrium (Option- “B”)
Note Given that the body is in equilibrium. If the sum is zero then the body will not rotate. So, option “B” is correct. If the sum is positive the body turns anticlockwise so option “C” is not correct. If negative then it turns clockwise so option “D” is also not correct.
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