Understanding Friction on Slopes: Choosing Between Cos and Sine Components

[Shovna]

The question is quite simple by i dnt seem to be getting it..
When a body is placed in a slope.
force mg is resolved into rectangular components..
the one perpendicular to the slope is taken as cos component n the one parallel is taken as sine component.
I don't understand whisch one to take as cos and which one as sine..generally x-axis is cos n y-axis is sine bt m unable to make out in a figure that which ione to take as x-axis n which one as y-axis??


pleasez help

 

[Doc Al]

When you want the x & y components of a vector F and the vector's angle is given with respect to the x-axis, then the x-component will be Fcosθ.

But when resolving weight into components parallel (usually called the x-axis) and perpendicular (y-axis) to the incline, realize that the angle of the incline is the angle that the weight makes with the y-axis, not the x-axis. So the trig functions end up getting reversed.

Read this: http://www.physicsclassroom.com/Class/vectors/u3l3e.cfm"

 

[seto6]

your F_g fully does not constrabute to friction... becasue the y- component is driving the object down.. or making it stationary