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Chapter 6: Q6MP (page 336)

Find the torque about the point (1, -2, 1) due to the forceF = 2 i - j + 3 kacting at the point ( 1, 1, -3)

Short Answer

Expert verified

The torque is found to be $τ = 5 i - 8 j - 6 k$ .

Step by step solution

01

Given Information.

It has been given that the point is (1, -2, 1) and ( 1, 1, -3) and the force is F = 2 i - j + 3 k

02

Definition of vector.

A quantity that has magnitude as well as direction is called a vector. It is typically denoted by an arrow in which the head determines the direction of the vector and the length determines it magnitude.

03

Use the formula for torque.

The torque is given by the formula.

$τ = r \times F$

Here is the vector pointing from the point of question to the point at which the force is acting.

$r = (1, 1, - 3) - (1, - 2, 1) = 3 j - 4 k$

Given $F = 2 i - j + 3 k$ , calculate as simple cross product.

$τ = r \times F = |\begin{matrix} i & j & k \\ 0 & 3 & - 4 \\ 2 & - 1 & 3 \end{matrix}| τ = ((3) (3) - (- 4) (- 1)) i - ((0) (3) - (- 4) (2)) j + ((0) (- 1) - (3) (2)) k τ = 5 i - 8 j - 6 k$

Hence, the torque is found to be $τ = 5 i - 8 j - 6 k$ .

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Most popular questions from this chapter

$\iint V \times ndσ$ over the closed surface of the tin can bounded by if $x^{2} + y^{2} = 9, z = 0, z = 5$ if $V = 2 x y i - y^{2} j + (z + x y) k$ .

Given that $B = curl A$ , use the divergence theorem to show that over any closed surface is zero.

Evaluate each of the following integrals in the easiest way you can.

$\int (y^{2} - x^{2}) d x + (2 x y + 3) d y$ ,along the xaxis from (0,0) and localid="1659182150932" $(\sqrt 5, 0)$ then along

a circular arc from $(\sqrt 5, 0) to (1, 2) .$

For a simple closed curve Cin the plane show by Green’s theorem that the area inclosed is

$A = \frac{1}{2} \oint_{C} (x d y - y d x)$

(a) Given $Φ = x^{2} - y^{2}$ , sketch on one graph the curves $Φ = 4, Φ = 1, Φ = 0, Φ = - 1, Φ = - 4$ . If $Φ$ is the electrostatic potential, the curves $Φ =$ const. are equipotential, and the electric field is given by $E = - \nabla Φ$ . If $Φ$ is temperature, the curves $Φ$ = const. are isothermals and $\nabla Φ$ is the temperature gradient; heat flows in the direction $- \nabla Φ$ .

(b) Find and draw on your sketch the vectors $- \nabla Φ$ at the points $(x, y) = (+ 1, + 1)$ , $(0, + 2), (+ 2, 0)$ ,. Then, remembering that $\nabla Φ$ is perpendicular to $Φ$ = const., sketch, without computation, several curves along which heat would flow [see (a)].

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