Reviewing Cross Product: Simplest Method Possible

In summary, the conversation is about someone asking for help with a physics problem and admitting that they did it in a complicated way. They hope someone can provide a simpler method and want a step by step solution to fully understand it. The person providing the solution explains the formula for torque and how to find the force using cross product and dot product. They also joke about a green stain on the final answer and discuss different ways to solve the problem.
  • #1
ineedhelpnow
651
0
HEY GUYS! (Wave)
ok so i have this question i did. and now I am reviewing for the test and i looked at how i did it and i did in the most complicated way ever. i don't FULLY understand chegg's method. so i hope someone can provide me with the SIMPLEST method possible. thank u! (Blush) (p.s. don't ask me what the green stain on my final answer is because i have no idea) i would really appreciate a step by step solution at one shot so i can fully understand it.

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$\left| \tau\right| = \left| r \right| \left| F \right| sin (\theta)$
 

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  • #2
ineedhelpnow said:
HEY GUYS! (Wave)
ok so i have this question i did. and now I am reviewing for the test and i looked at how i did it and i did in the most complicated way ever. i don't FULLY understand chegg's method. so i hope someone can provide me with the SIMPLEST method possible. thank u! (Blush) (p.s. don't ask me what the green stain on my final answer is because i have no idea) i would really appreciate a step by step solution at one shot so i can fully understand it.

Your result is correct! (Yes)

Here is how I would do it.

The formula for torque is:
$$\boldsymbol{\tau} = \mathbf r \times \mathbf F \tag 1$$

Normalizing the direction vector of the force, we have:
$$\mathbf{\hat F} = \frac 1 5 \begin{pmatrix}0\\3\\-4\end{pmatrix} \tag 2$$

Substituting the data of $\mathbf r$ and $(2)$ into $(1)$ we get:
$$\boldsymbol\tau
= \begin{pmatrix}0\\0.3\\0\end{pmatrix}
\times \frac 1 5 F \begin{pmatrix}0\\3\\-4\end{pmatrix}
=\frac{1}{50}F \begin{pmatrix}0\\3\\0\end{pmatrix}
\times \begin{pmatrix}0\\3\\-4\end{pmatrix}
=\frac{1}{50}F \begin{pmatrix}-12\\0\\0\end{pmatrix}
$$

So:
$$\tau=\frac{12}{50}F=100 \Rightarrow F = 417 \text{ N}$$

Btw, what is that $\color{Green}{\text{green stain}}$ on your answer?
Were you eating? (Wondering)
 
  • #3
ive never really seen it don't that way. is there an easy way to first solve for theta and then continue the rest of the problem.

lol i don't know. i think its a food stain. i was eating while doing it. i think its like oil or something (Giggle)
 
  • #4
now that i look back at it, i think it does make sense the way i did it. when i was doing it, i didnt get it much but now i realize that all i did was use the dot product as well as the cross product. i think i understand it more now because I am familiar with the equation $r \cdot v= \left| r \right| \left| v \right| cos (\theta)$
 
  • #5
ineedhelpnow said:
ive never really seen it don't that way. is there an easy way to first solve for theta and then continue the rest of the problem.

That's how your solution works.
First use the dot product to find the angle between the 2 vectors:
$$\mathbf r \cdot \mathbf v = r_x v_x + r_y v_y + r_z v_z$$
$$\mathbf r \cdot \mathbf v = |r|\cdot |v|\cdot \cos(\theta)$$
Then use the cross product to find the torque:
$$\tau = |\mathbf r \times \mathbf F| = |r|\cdot |F|\cdot \sin(\theta)$$

In my version I skipped those steps with the dot product, and went straight for the cross product.
 

FAQ: Reviewing Cross Product: Simplest Method Possible

What is a cross product?

A cross product is a mathematical operation that takes two vectors as input and produces a third vector that is perpendicular to both of the input vectors.

What is the simplest method for reviewing cross product?

The simplest method for reviewing cross product is to follow the steps of the cross product formula, which involves finding the determinant of a 3x3 matrix and using it to calculate the components of the resulting vector.

Why is the cross product important?

The cross product is important because it allows us to find a vector that is perpendicular to two other vectors, which is useful in many applications such as calculating the torque in a physical system or finding the direction of a magnetic field.

What are some common mistakes when calculating the cross product?

Some common mistakes when calculating the cross product include forgetting to take the order of the vectors into account, using the wrong formula, or making errors in calculating the determinant of the matrix.

What are some real-world examples of using the cross product?

The cross product has many real-world applications, such as calculating the moment of force in a lever system, determining the direction of a magnetic field in physics, or finding the normal vector to a surface in computer graphics.

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