Consider two inclined forces on the link. 100N at 60 ̊ counter clockwise direction and 450N...

Consider two inclined forces on the link. 100N at 60 ̊ counter clockwise direction and 450N at 30 ̊ clockwise direction. Determine the resultant force and its direction on the link using components method.

Expert Solution

100N can be resolved into 2 components. 100cos60 (along positive X axis) and 100sin60(along positive Y axis). 450N can be resolved into 450cos30 along positive X axis and 450sin30 along negative Y axis.

Resultant force in X direction Fx=100cos60+450cos30=50+389.711=439.711N (because these two x components are in same direction )

Resultant force in y direction

Fy=100sin60-450sin60=-303.109N (because 100sin60 is along positive y axis and 450sin60 is along negative y axis. Hence net force of these two is in negative y direction )

Resultant of Fx and Fy, F===534.06N (square of negative term is positive )

Let be the angle between F and Fx

Hence tan=Fy/Fx=-303.109/439.711=-0.6893

=- (That is F is at 34.58 degee below with x axis

genius_generous answered 2 years ago

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