Calculating the Direction of a Block's Travel from Multiple Forces

[MCATPhys]

Simple question: 20 kg block pushed from the east with 100 N, from the west with 20 N, and from south with 150 N. In what direction does the block travel.

net x = 100-20 = 80 towards west
net y = 150 towards north

tan (x) = 150/80
x = 61.9 degrees

For some reason, the book says the answer is "151.9 degrees measured counterclockwise from east" Can someone tell me what I did wrong please.

 

[CanIExplore]

Draw the vectors on a coordinate system and you'll probably immediately see where you've gone wrong. Let the north be in the positive y direction and east be in the positive x direction.

 

[MCATPhys]

Draw the vectors on a coordinate system and you'll probably immediately see where you've gone wrong. Let the north be in the positive y direction and east be in the positive x direction.

I did do that. The resultant x force is towards the left, so I drew it on the -x axis. the resultant y force is towards the north, so I drew it on the positive y axis.

 

[CanIExplore]

I did do that. The resultant x force is towards the left, so I drew it on the -x axis. the resultant y force is towards the north, so I drew it on the positive y axis.

Notice the total vector is in the second quadrant. You have to take the signs of the components into account. So you've got the magnitude of the angle right but that's for a vector in the first quadrant.

 

[MCATPhys]

I did all my calculations on the second quadrant.

So I drew a line on the -x axis that corresponds to the 80N, and at its tip I drew another vector pointing up corresponding to the 150N. So the resulting angle is in the second quadrant. The answer I got was 61.9 degrees north of west. In other words, 180-61.9 = 118.1 degrees from the first quadrant.

 

[EngineerHead]

+x direction is towards the east, -x direction is towards the west. The net x-force is towards the west (therefore negative direction/towards -x). When solving for the angle, it should be:

tan(\phi) = \frac{150}{-80}

You must visualize and be able to understand exactly in what direction these angles are relative to. Solving for this angle (equation above), you get -62\circ, which is measured "CCW" with respect to the westward horizontal (measured CCW - not CW - with respect to the westward/-x axis; this is because (-)CCW is actually (+)CW - MEANING: actual solved value (the negative value) is measured CCW from west axis, however if using the absolute value of this angle, it would be measured CW since you got rid of the negative). However, this very same angle measured CCW with respect to the the +x horizontal would have to be 180 - ans (62) = 118. This should be the correct answer, if the book says 152 degrees from the +x axis, then I believe it is wrong (rare).

Simple proof:
150>80, and therefore the angle measured CCW with respect to the +x axis should be <135 degrees. 118 matches this requirement, 152 does not.

 

[MCATPhys]

+x direction is towards the east, -x direction is towards the west. The net x-force is towards the west (therefore negative direction/towards -x). When solving for the angle, it should be:

tan(\phi) = \frac{150}{-80}

If I do that... I get -61.9 degrees, which points in the fourth quadrant...

 

[EngineerHead]

^ See edited post.

 

[MCATPhys]

I understand now... thanks so much