# 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°...

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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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.**

**Sources:**

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.

The force applied along the handle is 40 lb.

The angle of handle with ground is 30 deg.

The two components are to be determine, one in the forward direction i.e. along the ground and other perpendicular to the ground i.e. downwards.

The forward force is given by 40cos(30) = 34.64 lb.

The downward force is given by 40sin(30) = 20 lb.

**Thus the forward force is 34.64 lb **and** downward force is 20 lb.**

Since I can't draw the diagram on here. I will do my best to explain how the problem works.

1. You have the value of one angle (30 degrees) which means the other 2 angles must be 60 and 90. So we know we are dealing with an right angle. This means we can use trig functions and Pythagorean Theorem at some point.

2. The diagram should have vectors facing downward and forward to complete the triangle. So the vectors we have to define are opposite and adjacent to the 30 angle. This means that we have to use sin 30 and cos 30 in the equation.

3. I want to convert the pounds (lb) to newtons. The conversion goes like so:

40lbs x 4.45N/lbs= 178N

4. Lets figure out the values of the vectors

178N x cos 30 = 154.15N

178N x sin 30 = 89N

To check this answer you can do the Pythagorean Theorem

154.15^2+89^2=178^2

**Therefore the force she is exerting forward is 151.15N and the force she is exerting downward is 89N.**