Two workers pull horizontally on a heavy box, but one pulls twice as hard as the other. the larger pull is directed at 25.0 ∘ west of north, and the resultant of these two pulls is 590.0 n directly northward. use vector components to find the magnitude of each of these pulls. assume that the smaller pull has a component directed to the north and use vector components to find the direction of the smaller pull. assume that the smaller pull has a component directed to the north.

&=32.3°

Small pull=251.38N

Large pull=502.76N

Small pull is 32.3° north east

&=-32.3°

Small pull=416.58N

Large pull=923.17N

Small pull is 32.3°south east

Explanation: Let the small pull have a magnitude of F

Let the larger pull have a magnitude of 2F

On x-component:

Fcos&-2Fsin25=0

F(cos&-2sin25=0

F=0, we have,

Cos&-2sin25=0

&=Cos-1(2sin25)

&=Cos-1 0.8452

&=32.3° or-32.3°

X-component using &=32.3°

Fy+2Fy=Fsin32.3+2Fcos25=590N

F=590/(sin32.3+2cos25)

F=590/(0.5344+1.8126)

F=590/2.3470=251.38N

2F=2×251.38×2=502.76N

For&=-32.3°

Fsin(32.3)+2cos25=590N

F=590/(sin(-32.3) +2cos25)

F=590/(-0.5345+1.8126)

F=590/1.2782 = 461.59N

2F=923.17N

1 - wave length

2 - amplitude

3 - crest

4 - trough

explanation:

duh.

what was the answer?