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# components of forces help. watch

1. how do i approach this question?

2. Are you familiar with the dot product? If so, the component of force in a direction is the force dotted with the unit vector in that direction. And since the force and direction are at right angles...
3. all the forces given from a up parrallel to the hill on the block.

so Because F = MA

since there is no mention of whether it is moving IE is there a friction force we are going to assume that F is not acting. So

F = -Wcos(theta)

notice that when this happens there is no component of R that makes up the total force when resolved parrellel to the hill. this is because R is perpendicular to the direction in which you resolved the forces
4. (Original post by soup)
Are you familiar with the dot product? If so, the component of force in a direction is the force dotted with the unit vector in that direction. And since the force and direction are at right angles...
Ah i get it now, When resolving the component parallel is the horizontal (RCos(theta)) since the force and direction of R = 90 degrees, therefore RCos(90) = 0.

thanks soup
5. (Original post by CoolRunner)
Ah i get it now, When resolving the component parallel is the horizontal (RCos(theta)) since the force and direction of R = 90 degrees, therefore RCos(90) = 0.

thanks soup
Even if you just tilt your head a bit or the diagram, you should straight away notice how R is a straight perpendicular vertical line to the inclined surface. So it cannot have any horizontal component parallel to the inclined surface.

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