Problem 1 solution

(1) FBD of the entire frame (3 pts; −0.5 for no xy axes, −1 pt for each incorrect force)

MD=0 ⇒ (400 N)(0.1 m) − (Ay)(0.4 m)=0 ⇒ Ay=100 N (3 pts; −1pt for incorrect value)

Fx=0 ⇒ Dx= 0 N (1 pt)

Fy=0 ⇒ Ay + Dy − 400 N = 0 N

100 + Dy − 400 N = 0 N ⇒ Dy= 300 N (1 pt; − 0.5 pt for incorrect value)

(2) FBD of ABCD (4 pts; −0.5 for no xy axes, −1 pt for each incorrect force)

CE is a 2-force member (1 pt)

MB=0 ⇒ (Dy)(0.3 m) + (C sin 45o)(0.2 m) − (Ay)(0.1 m)=0

(300 N)(0.3 m) + (C sin 45o)(0.2 m) − (100 N)(0.1 m)= 0

C = − 565.68 N = − 566 N (4 pts; −1pt for incorrect moment of C, −1pt for incorrect value)

Fx=0 ⇒ Bx − C cos 45o = 0

Bx − (− 565.68 N) cos 45o = 0 ⇒ Bx = − 400 N (1.5 pt; − 0.5 pt for incorrect value)

Fy=0 ⇒ Ay + By + C sin 45o + Dy = 0 N

100 + By + (− 566 N) sin 45o + 300 N = 0 N ⇒ By = 0 N (1.5 pt; − 0.5 pt for incorrect value)

(− 0 pt for not showing units once, − 0.5 pt for not showing units twice, − 1 pt for not showing units 3 times or more).

Solution:

a) 𝐹 = −𝐹 sin 45° = −0.707 𝐹

𝐹 ′ = 𝐹 cos 45° = 0.707 𝐹

𝐹 = 𝐹 cos 20° = 0.707 𝐹 cos 20° = 0.664 𝐹

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𝐹 = 𝐹 cos 20° = 0.707 𝐹 sin 20° = 0.242 𝐹

𝐅𝐁 = { 0.664 𝒊 + 0.242 𝒋 − 0.707 𝒌} 𝐹 lb

The force vector FC is given as:

𝐅𝐂 = −0.664 F 𝒊

The resultant of forces, FB and FC is:

FR = FB + FC

⇒ 𝐅𝐑 = {0.664 𝒊 + 0.242 𝒋 − 0.707 𝒌} F + {−0.664 F 𝒊}

⇒ 𝐅𝐑 = {0.242 𝒋 − 0.707 𝒌} F lb

b) The position vector along the spring:

𝒓𝑶𝑨 = {0.45 𝒋 − 1.41 𝒌} m

The unit vector along the spring is:

𝒖𝑶𝑨 =𝒓𝑶𝑨

𝑟=

{0.45 𝒋 − 1.41 𝒌}

(0.45) + (−1.41)= 0.304 𝒋 − 0.953 𝒌

The magnitude of a component of FB parallel to the spring is:

F∥

= 𝑭𝑩 ⋅ 𝒖𝑶𝑨 = (0.664 𝐹 )(0) + (0.242 𝐹 )(0.304) + (−0.707 𝐹 )(−0.953)

= 0.747 𝐹 lb

The magnitude of a component of FB perpendicular to spring is:

F = F − F∥

= F − (0.747 F )

= 0.664 F lb

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Give full points if rAO and uAO used in place of rOA

and uOA

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Give full points if they directly write F

∥= 0 and

F = 0.664 F lb

The magnitude of a component of FC parallel to the spring is:

F∥

= 𝑭𝑪 ⋅ 𝒖𝑶𝑨 = (−0.664 𝐹 )(0) + (0)(0.304) + (0)(−0.953) = 0 lb

The magnitude of a component of FB perpendicular to spring is:

F = F − F∥

= (0.664 F ) − (0) = 0.664 F lb

c) Given: F = 20 lb and the undeformed length of the spring, l’ = 1.08 ft

For equilibrium, the magnitude of spring force = F

The deformed length of the spring, 𝑙 = OA = √0.45 + 1.41 = 1.48 ft

Let l’ be the undeformed length of the spring.

Then, deformation of the spring, x = l – l’ = (1.48 – 1.08) ft = 0.4 ft

F = k x = k (𝑙 − 𝑙 )

⇒ k = F

𝑙 – 𝑙’

⇒ k =.

= 50 lb/ft

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Final answer

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Problem #5 (20 points)

The forces act on the frame as shown. Determine:

(a) An equivalent force-couple system at A. (12) (b) The scalar component of the moment generated about AD (6) (c) The vector component of the moment generated about AD (2)

A) �⃗� = −400𝑖 + 200𝑗 + 500𝑗 − 600𝑘 + 300 cos(56°)𝑖 − 300 sin(56°) 𝑗

𝑭𝑹 = [ −𝟑𝟐. 𝟐 𝒊 + 𝟐𝟓𝟏𝒋 − 𝟔𝟎𝟎𝒌]𝑵

[4 pts] (2 pts correct force components, 2 for correct answer)

�⃗� =𝑖 𝑗 𝑘0 0 1

−400 0 0

+𝑖 𝑗 𝑘0 1 1.50 500 0

+𝑖 𝑗 𝑘

−0.75 1 1.50 0 −600

+𝑖 𝑗 𝑘

−0.75 1 1.5167.8 −248.7 0

�⃗� = −400𝑗 − 750𝑖 + (−600𝑖 − 450𝑗) + (373.1 𝑖 + 251.6 𝑗 + 18.87𝑘)

�⃗�𝑹𝑨 = [−𝟗𝟕𝟕𝒊 − 𝟓𝟗𝟖𝒋 + 𝟏𝟖. 𝟖𝒌]𝑵 ∗ 𝒎

[8 pts] (4 points for correct cross-products, okay to combine 300N and 600N, 2 pts for showing cross-product results. 2 pts for correct answer)

B) 𝜇 =⃗

|| ⃗ ||=

[ . . ]

√ . .= −0.384𝑖 + 0.512𝑗 + 0.768𝑘

𝑀 =−0.384 0.512 0.768

0 0 1−400 0 0

+−0.384 0.512 0.768

0 1 1.50 500 0

= −204.8 + 288 = 83.2 N*m

[6 pts] (2 points for unit vector, 3 points for correct cross products, 1 pt for correct answer)

C) �⃗� = 𝜇 ∗ 𝑀 = (83.2 𝑁 ∗ 𝑚) ∗ (−0.384𝑖 + 0.512𝑗 + 0.768𝑘)

�⃗�𝑨𝑫 = (−𝟑𝟏. 𝟗𝒊 + 𝟒𝟐. 𝟔𝒋 + 𝟔𝟑. 𝟗𝒌)𝑵 ∗ 𝒎

[2 pts] (1 point for unit vector times scalar moment, 1 point for correct answer)