Guest iblearner Posted May 1, 2016 Report Share Posted May 1, 2016 (edited) The correct answer is B. But why couldn't it be C or D? They all look similar to me lol Edited May 1, 2016 by iblearner Reply Link to post Share on other sites More sharing options...
ibstudent321 Posted May 1, 2016 Report Share Posted May 1, 2016 If you take the Normal vector and Friction vector to be on the y and x axis respectively, it means that Weight has both an x and y component. I think you need to resolve the Weight vector into it's components. Since it's at a constant velocity there is no resultant force and the "horizontal" and "vertical" components should be equal in both directions. I think that's why they gave you the graph paper background. C doesn't balance in the x direction D doesn't balance in the y direction Reply Link to post Share on other sites More sharing options...
Guest iblearner Posted May 1, 2016 Report Share Posted May 1, 2016 1 hour ago, ibstudent321 said: If you take the Normal vector and Friction vector to be on the y and x axis respectively, it means that Weight has both an x and y component. I think you need to resolve the Weight vector into it's components. Since it's at a constant velocity there is no resultant force and the "horizontal" and "vertical" components should be equal in both directions. I think that's why they gave you the graph paper background. C doesn't balance in the x direction D doesn't balance in the y direction I think I kinda understand now, thanks :)) Reply Link to post Share on other sites More sharing options...
kw0573 Posted May 1, 2016 Report Share Posted May 1, 2016 When the object is not accelerating in direction perpendicular to the ramp, then the normal force in this case must be equal in magnitude as the component of weight perpendicular to the ramp, not the other way around. For example if angle θ is the angle that the ramp makes with the horizontal, then the normal force of an object on a ramp is often expressed as mg*cos(θ), or magnitude of mg*cos(θ) ≤ magnitude of mg This is not always true depending on the setup of the question. Choice C shows gravitational force equal in magnitude as normal force, which is not true. Choice D shows gravitational force is smaller than normal force, which is also not true. Note that in Choice A, the component of weight I am referring to, is depicted instead of the correct downward weight. So that equal in magnitude and opposite direction is what we want, but we draw the weight downward. Reply Link to post Share on other sites More sharing options...
Guest iblearner Posted May 2, 2016 Report Share Posted May 2, 2016 19 hours ago, kw0573 said: When the object is not accelerating in direction perpendicular to the ramp, then the normal force in this case must be equal in magnitude as the component of weight perpendicular to the ramp, not the other way around. For example if angle θ is the angle that the ramp makes with the horizontal, then the normal force of an object on a ramp is often expressed as mg*cos(θ), or magnitude of mg*cos(θ) ≤ magnitude of mg This is not always true depending on the setup of the question. Choice C shows gravitational force equal in magnitude as normal force, which is not true. Choice D shows gravitational force is smaller than normal force, which is also not true. Note that in Choice A, the component of weight I am referring to, is depicted instead of the correct downward weight. So that equal in magnitude and opposite direction is what we want, but we draw the weight downward. I just read your reply, thanks so much ^^ Reply Link to post Share on other sites More sharing options...
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