Question

Consider three force vectors Fi with magni- tude 53 N and direction 116º, F2 with mag- nitude 57 N and direction 217°, and F3 with magnitude 71 N and direction 20°. All direc- tion angles 0 are measured from the positive x axis: counter-clockwise for 0 > 0 and clock- wise for 0 < 0. What is the magnitude of the resultant vec- tor || F ||, where F = Fi + F2 + 3 ? Answer in units of N. 004 (part 2 of 2) 10.0 points What is the direction of É as an angle between the limits of -180° and +180° from the positive x axis with counterclockwise as the positive angular direction? Answer in units of 005 10.0 points Consider the instantaneous velocity of a body. This velocity is always in the direction of 1. the least resistance at that instant. 2. the net force at that instant. 3. the motion at that instant.

Answer 1

Answer:

a. Fnet =37.67N

b. The direction = 133.4 from the x axis counter clockwise.

c. Option 2

Explanation:

Given that F1 is 53N at 116°, then it will be at a direction of 116-90=26° in the second quadrant.

Given that F2 is 57N at 116°, then it will be at a direction of 217-180=37° in the third quadrant..

Given that F1 is 71N at 20°, then it is in the first quadrant.

a. Fnet= F1+F2+F3

Fnet= -F1sin26i+F2cos26j-F2cos37i-F2sin37j+F3cos20i+F3sin20j

Fnet= 53sin26i+53cos26j-57cos37i-57sin37j+71cos20i+71sin20j

Resolving the vectors into x and y components.

Fnet= -2.04i+37.62j

Magnitude of the vector

Fnet= √((-2.04)^2+(37.62)^2)

Fnet= 37.67N

Fnet is approximately 38N.

b. Direction of the Fnet.

Angle=arctan(y/x)

Angle=arctan(-37.61/2.04)

Angle= -43.37°

The angle is in the negative x axis and positive y axis.

Then the direction becomes 180-43.37

Therefore, the direction of the net force is 133.37°.

c. The instantaneous velocity of a body is always in the direction of the net force at that instant. Option 2 is correct.

Fnet=ma

Fnet= mv/t

So the velocity is in the direction of the Fnet.