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24-351

HW 2 Solution

Due: 9/17/15

1 As illustrated below, a frame ABC is supported by a cable DBE that passes through a ring (B). If the tension in the cable is 385 N, what is (a) the force that it exerts on the support D and (b) the resultant force that it exerts on the ring?

Fx = 240 N Fy = –255 N Fz = 160 N [1/1.5/2]

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24-351

HW 2 Solution

Due: 9/17/15

[1/1.5/2] 2 A 50 lb load (C) is suspended by a cable that passes through a pulley (B) and is connected to a collar (A) that slides on a frictionless rod. Referring to the illustration below, determine the following: a) The force P necessary to maintain equilibrium when x = 15 in. b) The distance x required to maintain equilibrium when P = 48 lb?

20 in

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HW 2 Solution

24-351

Due: 9/17/15

[1/1.5/2]

[1/1.5/2]

3 A 800 N load is acting on the bracket shown on the right. What is the moment about MB point B. Make sure to indicate the direction of moment, i.e. clockwise or counterclockwise. Refering to the FBD below, the moment about point B is calculate as MB = rA/B × F where rA/B = –(0.2 m)i + (0.16 m)j F = (800 N)cos(π/3)i + (800 N)sin(π/3)j = (400 N)i + (693 N)j ⇒ MB = {–(0.2 m)i + (0.16 m)j} × {(400 N)i + (693 N)j} !

HW 2 Solution

24-351 = –(202.6 N⋅m)k ∴ MB = 203 N⋅ m clockwise

Due: 9/17/15

[2/3.5/4]

4 As illustrated below, three cables are tied to the bracket and subject to tension. Replace the exerted forces with an equivalent force-couple system at point A. Make sure to express the resultant force R and moment MA as vectors with components for each of the i, j, k bases.

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HW 2 Solution

24-351

Due: 9/17/15

FB = (300 N)i – (600 N)j + (200 N)k FC = (707 N)i – (707 N)k FD = (600 N)i + (1039 N)j F = FB + FC + FD = (1607 N)i + (439 N)j – (507 N)k rB/A = 0.075i + 0.050k rC/A = 0.075i – 0.050k rD/A = 0.100i – 0.100j (MA)B = rB/A × FB = (30 N⋅m)i – (45 N⋅m)k (MA)C = rC/A × FB = (17.68 N⋅m)j (MA)D = rD/A × FB = (163.9 N⋅m)k MA = (MA)B + (MA)C + (MA)D = (30 N⋅ m)i + (17.68 N⋅ m)j + (118.9 N⋅ m)k

[3/5.5/6]

5 Use the triple scalar and vector product identities to prove the following: (A×B)⋅(C×D) = (A⋅C)(B⋅D) – (A⋅D)(B⋅C) According to the triple vector product identity: (C×D)×A = (A⋅C)D – (A⋅D)C Next, define M = C×D take the scalar product of both sides with B: (M×A)⋅B = (A⋅C)(D⋅B) – (A⋅D)(C⋅B) According to the triple scalar product identity: (M×A)⋅B = M⋅(A×B) Lastly, noting that the scalar product is commutative, i.e. X⋅Y = Y⋅X, (A×B)⋅M = (A⋅C)(B⋅D) – (A⋅D)(B⋅C) ∴ (A× × B)⋅⋅ (C× D) = (A⋅⋅ C)(B⋅ D) – (A⋅ D)(B⋅ C) [1/1.5/2]

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