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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 蟿=5i-8j-6k.

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.

蟿=r脳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)=3j-4k

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

蟿=r脳F=ijk03-42-13蟿=((3)(3)-(-4)(-1))i-((0)(3)-(-4)(2))j+((0)(-1)-(3)(2))k蟿=5i-8j-6k

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

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Most popular questions from this chapter

Givenu=xy+yz+zsinx, find

(a) 鈭�uat(0,1,2);

(b) The directional derivative of (0,1,2) at in the direction 2i+2j-k;

(c) The equations of the tangent plane and of the normal line to the level surface u=2at(0,1,2);

(d) a unit vector in the direction of most rapid increase of u at(0,1,2)

Write out the equations corresponding to (9.3) and (9.4) for Q=2Xbetween points 3 and 4 in Figure 9.2, and add them to get (9.6).

Do case (b) of Example 1above.

Find the derivative of xy2+yz at (1,1,2) in the direction of the vector 2i-j+2k.

Use Green鈥檚 theorem (Section 9) to do Problem 8.2.
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