course Phy 231
The equilibrant of a force is the force which is equal and opposite to that force. If two forces are equal and opposite, their x and y components are also equal, but the x and y components of the force are opposite in sign to those of the equilibrant. The x and y components of a force are 2 Newtons and 3 Newtons repectively.
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What are the magnitude of this force and what angle does it make as measured counterclockwise from the positive x axis?
2^2 + 3^2 = c^2
c =~ 3.6 N at an angle of 3.6cos(x) = 2, and x = 56.3 deg
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What are the components of the equilibrant force?
-2N for x, and -3N for y-comp.
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What angle does the equilibrant force make as measured counterclockwise from the positive x axis?
if it's equal and opposite to the first vector, it'd be 180 degrees rotated, so 180 + 56.3 = 236.3 deg.
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Sketch a vector representing a 10 Newton force which acts vertically downward.
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Position an x-y coordinate plane so that the initial point of your vector is at the origin, and the angle of the vector as measured counterclockwise from the positive x axis is 250 degrees. This will require that you 'rotate' the x-y coordinate plane from its traditional horizontal-vertical orientation.
answer/question/discussion:
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What are the x and y components of the equilibrant of the force?
answer/question/discussion:
10N (cos 250) = -3.4 N for x
10N (sin 250) = -9.4 N for y
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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.
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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:
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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:
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&&&&& If the object is moving in the direction 120 deg, it's moving from the origin into the second quadrant, and if the force is in the direction 270 deg, it's directed from the origin along the y axis (or the third/fourth quadrant line), so if you put them together, and you were standing on the vector in the second quadrant, the force would be going in a downward direction, and you'd be looking in a north-westerly direction, so it'd be coming toward you.
Will the object speed up, slow down or maintain a constant speed?
answer/question/discussion: &&&&&&&&&Since the force is opposing the line of motion, I'd imagine it would slow down after receiving the force, although, depending on the force, it might go faster in the opposite direction.
If the force acts for a long enough time, it will eventually go faster in the opposite direction. The short-term effect, as you understand, will be to slow it.
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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: &&&&&Well, since the 270 deg force is perpendicular to the x-axis, it would be perpendicular to the x-component of the 120 deg vector. So it would be perpendicular to the line of motion on the left, but the force would be exerted from the right of the object (object is moving northwest, force is exerted from north to south). So, since it's pointed at 270 degrees (i.e., if you moved the vector that's downward along the negative y-axis without changing its angle, you would have to put it above the vecter in the second quadrant for it to be exerted on the object).
Will the object veer to the right, to the left or maintain straight-line motion?
answer/question/discussion:
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&&&&If it gets hit moving in a north-westerly direction at 120 deg from a force moving in a southern 270 deg direction, I imagine, depending on the force, it would move anywhere from 120 to 270 deg after the force is exerted.
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:
&&&&&&&&&Well, the line of motion is 120 deg, and perpendicular is 30 deg or 210 deg, but I don't really know what to do next. Don't you have to have an actual value and not just the angles to determine which one is greater in magnitude?
Good work, but you need some clarification on the second question.
You should resubmit the second of these cq's, using the form from which it came. When you do so, the question is automatically labeled and I can quickly refer you to the appropriate link for a further discussion (which in this case addresses your questions). As it is I would either have to repeat that discussion or search out this question by trial and error in order to give you the reference.