tn

Fy

x

A

P

O

α

β

r

Figure 1: Problem 1

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���������� O A

B

b

h F

Figure 2: Problem 2

1. The control rod AP exerts a force F on thesector as shown. Determine both the x-y andthe n-t components of the force.Solution. F can be expressed asF = −F (sin βi+cos βj). Unit vectors, en andet, are attached to the n-t coordinates. en andet are described as en = − cos αi− sin αj andet = sin αi−cos αj, respectively. Substitutingi = − cos αen + sin αet and j = − sin αen −cos αet into the equation of F gives F =F [(cos α sin β + sin α cos β)en − (sin α sin β − cos α cos β)et] =F [sin (α + β)en + cos (α + β)et]. Thus,the x-y and n-t components of the forceare Fx = −F sin β, Fy = −F cos β,Fn = F sin (α + β), and Ft = F cos (α + β).

2. The force of magnitude F acts along the edgeof the triangular plate. Determine the momentof F about point O.Solution. We attach unit vectors i and j tothe horizontal and vertical direction. The forceF can be expressed as

F = −F

√b2 + h2

(bi + hj) (1)

y

Ox

A

B

5 m

12 kN

304 m

Figure 3: Problem 3

The moment about O is MO = −hj × F .

MO = −hj ×

[

−F

√b2 + h2

(bi + hj)

]

= −Fbh

√b2 + h2

k (2)

3. Replace the 12-kN force acting at point A bya force-couple system at (a) point O and (b)point B.Solution.

(a) We apply two equal and opposite 12-kN forces at O and identify the coupleMO = −4i × (12)(cos 30◦i + sin 30◦j) =−24k kN · m. Thus, the original force isequivalent to the 12-kN force at O andthe 24 kN · m couple.

(b) We apply two equal and opposite 12-kNforces at B and identify the couple MB =(−4i + 5j) × (12)(cos 30◦i + sin 30◦j) =−76.0k kN · m. Thus, the original forceis equivalent to the 12-kN force at B andthe 76.0 kN · m couple.

1