Asked by Sandara
Gravity is pulling downward on a 20.0kg block resting on a 15 degree incline. What is the magnitude of the force component parallel to the surface (hint: how to turn 20.0kg into force)?
Answers
Answered by
bobpursley
parallel down=m*g*cosTheta
Answered by
bobpursley
parallel down=m*g*SinTheta
correction here, sine, not cosine.
the friction on the plane down the plane is mu*mg*CosTheta
correction here, sine, not cosine.
the friction on the plane down the plane is mu*mg*CosTheta
Answered by
Sandara
Oh wow thanks! I was wondering if you could give me an explanation as to how you found the solution if that isn't too much to ask
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