How much force is she exerting forward and downward? Kristin is exerting a force of 40 lbs along the handle of a lawn mower as she pushes it to cut the grass. The mower handle makes an angle of 30° with the ground. I have drawn a model of this situation but can not come up with an answer. Please help! Applications with Vectors

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The force of 40lbs is along the handle. The question is asking us to resolve the force into x and y components (horizontal and vertical).

So we can draw a triangle where the hypotenuse is 40 lbs "long", and makes an angle of 30 degr with the horizontal

We can use...

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The force of 40lbs is along the handle. The question is asking us to resolve the force into x and y components (horizontal and vertical).

So we can draw a triangle where the hypotenuse is 40 lbs "long", and makes an angle of 30 degr with the horizontal

We can use trigonometric functions to find the other two sides of the triangle which are the vertical and horizontal components. To find the length of the side along the ground (the horizontal arm) use the definition of cosine: CosA=adjacent/hypotenuse. Here A=30 and hypotenuse=40, so adjacent=40*cos(30)=34.6.

To find the length of the vertical arm of the triangle use sinA=opposite/hypotenuse or sin30=opposite/40. This gives opposite=40*sin30=20.

The force has thus been resolved into its x and y components.

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You should use the process called vector's decomposition to find the components of the vector along x and y axis.

The problem provides the information that a force of 40 lbs is exerted along a handle and the handle's direction is of `30^o`  above the ground, that represents x axis, hence,the force vector direction if of `30^o`  above x axis

You need to decompose the force vector  using trigonometric definition of sine and cosine such that:

`cos 30^o = F_x/F =gt F_x = F*cos 30^o `

`F_x = 40*sqrt3/2 =gt F_x = 20sqrt3 lbs`

`sin 30^o = F_y/F =gt F_y = F*sin 30^o `

`F_y = 40*1/2 =gt F_y = 20 lbs`

Hence, evaluating the magnitudes of force vector components yields `F_x = 20sqrt3 lbs ` and  `F_y = 20 lbs` , where `F_x`  acts forward and `F_y`  acts downward, perpendicular to the ground.

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