Phy 231
Your 'cq_1_19.3' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** **
An object moving in the direction 120 degrees (as measured counterclockwise to the positive x axis) encounters a net force whose direction is 270 degrees.
• Sketch the force and its component along the line of motion, as well as its component perpendicular to the line of motion.
answer/question/discussion: ->->->->->->->->->->->-> :
The component along the line of motion moves straight down, parallel to the y axis. The component perpendicular to the line of motion is 0 because the angle is straight down (270 degrees). The object’s x component is -.5 and the y component is .866.
The line of motion is at 120 degrees, not along either axis.
#$&*
• Suppose you are facing in the direction of motion. Do you perceive the component of the force along the line of motion to be forward or backward? It this component in the direction of motion or opposite to the direction of motion?
answer/question/discussion: ->->->->->->->->->->->-> :
The force would seem to be moving backwards because it is going in sort of the opposite direction of the original motion. The angle makes the force go past and behind the observer.
#$&*
• Will the object speed up, slow down or maintain a constant speed?
answer/question/discussion: ->->->->->->->->->->->-> :
The object will probably slow down when it is acted on by the force. However, you would need to know how strong the force it to say for sure. If the force were strong enough it could make the object go faster.
any force at 270 degrees will tend to slow an object moving in the 120 degree direction, and will also push it to the left
#$&*
• If you are facing in the direction of motion, then the line perpendicular to the direction of motion will run to your right and to your left. Is the component of the force perpendicular to the line of motion directed to the right or to the left?
answer/question/discussion: ->->->->->->->->->->->-> :
The line of force perpendicular to the line of motion will be directed to the right of the observer.
#$&*
• Will the object veer to the right, to the left or maintain straight-line motion?
answer/question/discussion: ->->->->->->->->->->->-> :
The object will veer to the left because the y component of the force is greater than the x component, which will push the object to the left of its original path.
#$&*
• Which is greater in magnitude, the component of the force along the line of motion or the component perpendicular to the line of motion?
answer/question/discussion: ->->->->->->->->->->->-> :
The component along the line of motion (y component) has the greater magnitude.
#$&*
** **
30 minutes
** **
Check the link to be sure you have the right idea about the orientation of the various quantities. You should probably submit a revision, using &&&& for insertions.
&#See any notes I might have inserted into your document. If there are no notes, this does not mean that your solution is completely correct.
Then please compare your solutions with the expanded discussion at the link
Solution
Self-critique your solutions, if this is necessary, according to the usual criteria. Insert any revisions, questions, etc. into a copy of this posted document. Mark any insertions with &&&& so they can be easily identified.If your solution is completely consistent with the given solution, you need do nothing further with this problem. &#