Aeromechanics I-Honors - Description Of The Above Lectures

Homework Eleven Solution Exercise 1 Given: 1. uniform disk of weight W = 75 N with a rope attached to a wall. 2. friction between disk and ground: µ = 0.5. 3. rope with tension T at top. Find: the maximum value of T . Figure 1: Exercise 1 - problem setup. Solution: Draw a free body diagram (FBD): X F̄ = 0̄ : ĥ : −FW + T + Ff = 0 → FW = T + Ff ≥ 0 (rope) ĝ : W − N = 0 → W = N X M̄ O = 0̄ : (3rd component) (1) (2) (3) (4) ĥ : T r + (−Ff )r = 0 → T = Ff → (T ≥ 0 s.t. FW ≥ 0) 1 (5) Figure 2: Exercise 1 - FBD. We have friction between the disk and the ground: |Ff | ≤ µ|N | Eq.s (3) and (5) imply that T ≤ µ · W , which implies that Tmax = µW = (0.5)(75 N ) Tmax = 37.5 N 2 Exercise 2 Given: Scale of 6 lb to 1 inch. Find: Resultant and moment resultant about point O for this force system. Solution: F̄ = F¯1 + F̄2 + F̄3 + F̄4 → resultant expression. (6) √ p √ 6 lb Length of F̄1 and F̄2 = (5”)2 + (7”)2 = 74” = 8.6023 ⇒ 1 inch × 74” = 51.6138 lb. X Length of F̄3 and F̄4 = 4” ⇒ 6 lb 1 inch × 4” = 24 lb. We must now find unit vectors to express the force magnitudes along: 7 5 −7ẑ + 5ŷ √ = − √ ẑ + √ ŷ 74 74 74 7ẑ − 5ŷ 7 5 F̄2 ⇒ û2 = √ = √ ẑ − √ ŷ 74 74 74 F̄3 ⇒ û3 = 4x̂ = x̂ F̄4 ⇒ û4 = −4x̂ = −x̂ F̄1 ⇒ û1 = P so, the resultant F¯ = F1 û1 + F2 û2 + F3 û3 + F4 û4 is evaluated as X √  7 √  5  5  7 F̄ = 6 74 − √ ẑ + √ ŷ + 6 74 √ ẑ − √ ŷ + 24x̂ + 24(−x̂). 74 74 74 74 P Therefore F̄ = 0̄ lbs . The moment resultant is X M̄ O = N X i=1 3 r̄i × F̄i , (7) (8) (9) (10) (11) (12) where r̄i is a vector from O to a point on the line of action of F̄i . X M̄ O = r̄1 × F̄1 + r̄2 × F̄2 + r̄3 × F̄3 + r̄4 × F̄4 . Therefore, r̄1 = r̄OA = 4x̂ + 7ẑ r̄2 = r̄OB = 5ŷ r̄3 = 0̄ (because O is on the line of action of F3 ) r̄4 = r̄OC = 7ẑ. These equations allow us to evaluate the moment resultant equation as   X √  7 √  7 5  5  M̄ O = 4x̂ + 7ẑ × 6 74 − √ ẑ + √ ŷ + (5ŷ) × 6 74 √ ẑ − √ ŷ 74 74 74 74 + 0 + 7ẑ × −24x̂. X M̄ O = (4x̂ + 7ẑ) × (−42ẑ + 30ŷ) + (5ŷ) × (42ẑ − 30ŷ) − 168ŷ Thus, the resultant moment is P M̄ O = 120 lb·in ẑ . 4 (13) Exercise 3 Given: The force system as shown in the figure below. Find: A single force which is equivalent to this force system. Obtain the point of application of the single equivalent force. Solution: X F̄ = F̄1 + F¯2 + F̄3 + F̄4 = −80 lb ẑ + 90 lb ẑ − 120 lb ẑ − 100 lb ẑ F̄ = −210 lb ẑ . X M̄ 1O = 0 × F¯1 + r̄2 × F̄2 + r̄3 × F̄3 + r̄4 × F̄4   = 0 + (3ŷ × 90 lb ẑ) + ((−x̂ + 5ŷ) × −120 lb ẑ) + (−2x̂ + 1ŷ) × −100 lb ẑ = 270 x̂ − 120 ŷ − 600 x̂ − 200 ŷ − 100 x̂ = −430 x̂ − 320 ŷ ft · lb. X X O M̄ 2O = r̄ × M̄1 , where r̄ is from O to the point of application. F̄ = X F̄ = −430 x̂ − 320 ŷ ft · lb r̄ × X Let r̄ = ax̂ + bŷ ft. Substituting in the relevant quantities gives X M̄ 2O = (ax̂ + bŷ) × −210 lb ẑ = −430 x̂ − 320 ŷ ft · lb a · 210 lb ŷ − b · 210 lb x̂ = −430 x̂ − 320 ŷ 5 32 ŷ : a · 210 = −320 ⇒ a = − 21 43 x̂ : −b · 210 = −430 ⇒ b = . 21 32 x̂ + Thus, r̄ = − 21 43 ŷ 21 . To check the answer, we note that the equation must be satisfied r̄ × −210 lb ẑ = −430 x̂ − 320 ŷ ft · lb  32 43  − x̂ + ŷ × −210 lb ẑ ⇒ 21 21 = −430 x̂ − 320 ŷ ft · lb 6 Exercise 4 Given: The force system as shown in the figure below. Figure 3: Exercise 4 - Problem Setup. Find: A single force which is equivalent to this force system. Obtain the point of application of the single equivalent force (if applicable). Solution: Note the location of point P in Fig. 4. X F̄ = F̄1 + F̄2 + F¯3 + F¯4     = −50 N x̂ + 60 N ŷ + 120 N cos θ x̂ − sin θ ŷ + 100 N cos (45◦ ) x̂ + sin (45◦ ) ŷ Letting θ = tan−1 (2/1) = 63.43◦ , the resultant force is found to be P F̄ = 74.3856 N x̂ + 23.3841 N ŷ . For the moment resultant, we have X M̄ 1P = r̄1 × F̄1 + r̄2 × F̄2 + r̄3 × F̄3 + r̄4 × F̄4       = − 2 m ŷ × −50 N x̂ + − 2 m x̂ × 60 N ŷ + − 1 m x̂ × 100 N sin 45◦ ŷ = −100 Nm ẑ − 120 Nm ẑ − 70.7107 Nm ẑ = −290.7107 Nm ẑ. Note that X M̄2P = r̄ × X F̄ = −290.7107 Nm ẑ, 7 Figure 4: Exercise 4 - Grid in greater detail. and to find the point of application, we may use the equation P P F̄ × M̄1P r̄ = 2 P F F̄ × −290.7107 Nm ẑ = F2  P 74.3856 N x̂ + 23.3841 N ŷ × −290.7107 Nm ẑ = k74.3856 N x̂ + 23.3841 N ŷk2 (−6798.008 x̂ + 21624.670 ŷ) N2 m = (77.975)2 N2 r̄ = −1.118 x̂ + 3.557 ŷ m . To check the answer, we may use X     F̄ = − 1.118 m x̂ + 3.557 m ŷ × 74.3856 N x̂ + 23.3841 N ŷ r̄ × = −290.733 Nm ẑ. 8 Exercise 5 Given: The force system as shown in the figure below. Figure 5: Exercise 5 - Problem setup. Find: A single force which is equivalent to this force system. Obtain the point of application of the single equivalent force (if applicable). Solution: Figure 6: Exercise 5 - System in greater detail. To find the single force, we have X   F̄ = F̄1 + F̄2 + F̄3 = −50 lb ŷ − 60 lb x̂ + (70 lb) cos θ x̂ + sin θ ŷ . Noting that θ = 0.6435 rad, this evaluates to P¯ F = −4x̂ − 8ŷ . 9 Note the location of point P in Figure 6. To find the moment resultant, we have X     M̄ P = − 4 ft x̂ × −50 lb ŷ + 4 ft ŷ × −60 lb x̂ = 440 ẑ. To find the point of application r̄, we use the equation P P F × MP P 2 . r̄ = k Fk √ Note that k−4x̂ − 8ŷk = 80. Plugging in the force and moment resultants gives     − 4x̂ − 8ŷ × 440 ẑ √ = −44 ft x̂ + 22 ft ŷ . r̄ = ( 80)2 10