Question

Tom is climbing a 3.0-m-long ladder that leans against a vertical wall, contacting the wall 2.5 m above the ground. His weight of 480 N is a vector pointing vertically downward. (Weight is measured in newtons, abbreviated N.) Part A What is the magnitude of the component of Tom's weight parallel to the ladder? Express your answer with the appropriate units.

Answer 1

The magnitude of the component of Tom's weight parallel to the ladder can be found by using trigonometry. It is approximately 202.5 N.

Step-by-step explanation:

The magnitude of the component of Tom's weight parallel to the ladder can be found by using trigonometry. Since the ladder makes an angle of 25 degrees with the vertical wall, we can use the equation W|| = w sin(θ), where θ is the angle and w is the weight of Tom. So, the magnitude of the component of Tom's weight parallel to the ladder is:

W|| = 480 N * sin(25°)

Using a calculator, we find that the magnitude of the component of Tom's weight parallel to the ladder is approximately 202.5 N.

Answer 2

`W_(x)=400N`

Step-by-step explanation:

The very first thing we need to do is to draw a diagram that represents the situation. (See picture attached).

As you may see in the picture, we can incline the coordinate axis so the x-axis matches the ladder. When doing so, we can see the parallel component of Tom's weight will be opposite to the angle θ. So we can use the function sin to find the parallel component to the ladder for Tom's weight.

Also, notice that the distance between the base of the wall and the top of the ladder will also be opposite to the angle θ, while the length of the ladder will be the hypotenuse of the right triangle composed by the wall, the floor and the ladder.

we know that:

`sin(\theta)=(opposite)/(hypotenuse)`

so with the known data we can say that:

`sin(\theta)=(2.5m)/(3.0m)`

we can cancel the meters, leaving us just with:

`sin(\theta)=(2.5)/(3.0)`

Now, regarding the parallel component of Tom's weight, we now know that it will be found like this:

`sin(\theta)=(W_(x))/(W)`

when solving for
`W_(x)`, we get:

`W_(x)=Wsin(\theta)`

we can now substitute
`sin(\theta)` and W

so we get:

`W_(x)=(480N)((2.5)/(3.0))`

When calculating that, we get that the parallel component of Tom's weight is:

`W_(x)=400N`