Force 𝐅 acts perpendicular to the inclined plane. Determine the moment produced by 𝐅 about point

Force 𝐅 acts perpendicular to the inclined plane. Determine...

Key Concepts

Force

A force is a vector quantity that has both magnitude and direction. In mechanics, it is one of the fundamental influences that can cause objects to accelerate or change motion. Forces can be applied in various directions, and their analysis is key to understanding the overall behavior of a system.

Moment

The moment, or torque, represents the tendency of a force to cause rotation about a specific point or axis. It is mathematically defined as the cross product of the position vector (also known as the lever arm) and the force vector, and its magnitude depends on both the force applied and the perpendicular distance from the point of rotation.

Cross Product

The cross product is an operation on two vectors in three-dimensional space that produces a third vector perpendicular to the plane containing the first two. It is used in calculating moments because it naturally incorporates the perpendicular distance between the line of action of the force and the point about which the moment is taken.

Cartesian Vector Representation

A Cartesian vector representation expresses a vector in terms of its components along orthogonal x, y, and z axes. This form is particularly useful when performing vector operations, such as addition or the cross product, because it provides a clear and systematic framework for calculations.

Lever Arm

The lever arm is the position vector extending from the point about which the moment is calculated to the point where the force is applied. Its length and direction are crucial in determining the magnitude and direction of the moment produced by a force.

Transcript

00:01 Hello everyone here we have to calculate moment of force about b point due to force f is called to 400 muter applied perpendicular to incline plane now see here we will write the coordinates of the point a b c so a point having the coordinates 0 0 3 meter b b b b b b having the coordinates the coordinates 3 -0 -0 meter and c -point having the coordinates 040 meter we will find the position vector of ac so this could be written as 0 minus 0 i cap plus 4 minus 0 just a moment i have written here wrongly sorry 4 minus 0 j cap plus 0 minus 3 k cap plus 0 minus 3 k so all these values are in meter so this vector can be written as position vector ac can be written as after solving it you may write 4 j cap minus 3 k cap meter i am marking this equation to be one now fine position vector vc so this value will be 0 minus 3 i cap 4 minus 0 j cap plus 0 minus 0 k cap so position vector vc will be 4 j cap minus 3 k cap meter now since the force is acting perpendicular to inclined plane so we required a unit vector of force perpendicular to inclined plane uf we required uf vector which is unit vector perpendicular to inclined plane for that we will find a vector perpendicular to inclined plane and that will be r ac cross r bc vector now solve it using the determinant so this value will be 0 4 minus 3 i am substituting the value of position vector here minus 3 40 on solving this value will be 12 i cap plus 9 j cap plus 12 k cap meter square now unit vector perpendicular to inclined plane that is in the direction of force vector b divided by its magnitude so 12 i cam plus 9 j cap plus 12 k cap devised by its magnitude 12 square 9 kai square on solving this unit vector will be i am correcting here unit vector for unit vector cap.

05:01 So between the value unit vector you will get 0 .624 i cap plus 0 .469 j cap plus 0 .629 j cap plus 0 .624 5 k cap unit will be in meter now write the now we are writing the force in vector format it would be magnitude of force into its unit vector magnitude of force is 400 newton and unit direction of unit vector of force is 0 .6 to 4 .5 i cap plus 0 .469 g cap plus 0 .625 k cap on substituting the value and simplifying we will get the force to be 249 .9 i cap plus 0 .469 sorry i have written here wrongly i am doing the correction please pay attention this value will be 187 .4 newton along j -cap and 249 along k -cap newton.

07:32 Now we are calculating moment of force about b point, so cross -product of position vector vc and force solve it with the help of determinant and we will obtain the answer.

07:53 Let us see here how to solve.

07:56 First we write the position vector value minus 3, 4, 0...

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