# The chisel exerts a force of 20 lb 1

The chisel exerts a force of 20 lb on the wood dowel rod which is turning in a lathe. Resolve this force into components acting (a) along the n and t axes and (b) along the x and y axes. Image from: Hibbeler, R. C., S. C. Fan, Kai Beng. Yap, and Peter Schiavone. Statics: Mechanics for Engineers. Singapore: Pearson, 2013.

#### Solution:

We will first draw the components of the 20 lb force along the n and t axes. Using the sine and cosine functions, we can figure out the values $F_n$ and $F_t$. Remember that sine is opposite over hypotneuse, while cosine is adjacent over hypotneuse.

$\cos45^0\,=\,\dfrac{-F_n}{20}$

$F_n\,=\,-14.1$ lb

(Note how the value is negative, this is because force $F_n$ is heading towards the negative n-axis)

$\sin45^0\,=\,\dfrac{F_t}{20}$

$F_t\,=\,14.1$ lb

Now, we can focus on part (b) of the question. This time, we will draw our vector components along the x and y axes. (Notice how we extended the 20 lb force. This visual representation makes it easier for us to calculate the values required)

The $15^0$ angle can be found by realizing that if we extend the x-axis as shown in our diagram, it creates a semi-circle, which has an angle of $180^0$. Our force therefore must be applied at an angle of $15^0$ because $(45+60+30+60)-180^0=15^0$

As before, we can use sine and cosine functions to figure out the components along the x and y-axes.

$\cos15^0\,=\,\dfrac{F_x}{20}$

$F_x\,=\,19.3$ lb

$\sin15^0\,=\,\dfrac{F_y}{20}$

$F_y\,=\,5.18$ lb

a)

$F_n\,=\,-14.1$ lb

$F_t\,=\,14.1$ lb

b)

$F_x\,=\,19.3$ lb

$F_y\,=\,5.18$ lb

## One thought on “The chisel exerts a force of 20 lb”

• rao junaid iqbal

good