Question

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.

Answer 1

  The smaller pull is "F" and the larger pull is "2F". 
The east-west components of F and 2F cancel. 
The north components add to 460 N. 
2Fcos25 + FcosΘ = 460 
2Fsin25 - FsinΘ = 0 
where Θ is measured cw from +y axis (east of north) 
In the second equation, F cancels, leaving 
2sin25 = sinΘ = 0.845 → Θ = 58º East of North 
Using the first equation, then, 
2Fcos25 + Fcos58 = 1.813F + 0.534F = 2.347F = 460 
F = 196 N 
2F = 392 N

Answer 2

Answer:&=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