p170-w3.1-sols

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University of British Columbia *

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170

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Mechanical Engineering

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Oct 30, 2023

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PHYS 170 Worksheet 3.1 Solutions 1. Determine the greatest force F that can be applied if the cables are rated for 500 lb. ! A = 0 0 6 ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ! B = 3 − 2 0 ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ! C = − 3 − 1 0 ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ! D = − 2 3 0 ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ! r AB = ! B − ! A = 3 − 2 − 6 ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ! r AC = ! C − ! A = − 3 − 1 − 6 ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ! r AD = ! D − ! A = − 2 3 − 6 ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ L AB = 3 2 + 2 2 + 6 2 = 7.0000 L AC = 3 2 + 1 2 + 6 2 = 6.7823 L AD = 2 2 + 3 2 + 6 2 = 7.0000 Assume a vertical force of 1 pound: ! F = 0 0 + 1 ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ The equilibrium equation in vector form is ! r AB L AB T B + ! r AC L AC T C + ! r AD L AD T D + ! F = 0 Expand the vectors and move ! F : 3 − 2 − 6 ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ T B L AB + − 3 − 1 − 6 ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ T C L AC + − 2 3 − 6 ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ T D L AD = 0 0 − 1 ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ The augmented matrix is T B L AB T C L AC T D L AD 3 − 3 − 2 0 − 2 − 1 3 0 − 6 − 6 − 6 − 1 ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ Reduced row-echelon form: 1 0 0 0.07333 0 1 0 0.03333 0 0 1 0.06000 ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ The last column is the T L values. T B = 0.07333 ⋅ L AB = 0.5133lb T C = 0.03333 ⋅ L AC = 0.2261 lb T B = 0.06000 ⋅ L AD = 0.4200 lb Cable B has the highest tension of 0.5133lb per pound of force F, so it will break first So F max = 500 0.5133 = 974.1 lb.

2. The 40 kg block stretches the two springs to the lengths shown. Determine the lengths of the springs when the weight is removed. Note that the force exerted by a spring is is the change from the unstretched length. F = k Δ s , where Δ s 1.5 kN/m 1.2 kN/m θ AC θ AB We need the stretched lengths no matter how we do the problem. L AC = 0.6 2 + 0.5 2 = 0.7810 L AB = 0.4 2 + 0.5 2 = 0.6403 The weight force is W = mg = 40 ⋅ 9.81 = 392.4 We can get the force components from ratios of the sides of the right triangles. We write equations for x and y force balance, solve the simpler one symbolically, and plug into the other. x : − 0.6 L AC F AC + 0.4 L AB F AB = 0 → F AB = 0.6 L AC L AB 0.4 F AC y : 0.5 L AC F AC + 0.5 L AB F AB − W = 0 → 0.5 L AC F AC + 0.5 L AB 0.6 L AC L AB 0.4 F AC ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = W F AC = W 0.5 L AC + 0.5 ⋅ 0.6 0.4 ⋅ L AB = 392.4 0.5 0.7810 + 0.5 ⋅ 0.6 0.4 ⋅ 0.7810 = 245.2 N Then plug that result into the other equation. F AB = 0.6 L AC L AB 0.4 F AC = 0.6 0.7810 0.6403 0.4 ⋅ 245.2 = 301.5 N Now find the spring stretches, and subtract from the stretched lengths. Δ x AC = F AC k AC = 245.2 1200 = 0.2043m Δ x AB = F AB k AB = 301.5 1500 = 0.2010 m L AC after = 0.7810 − 0.2043 = 0.5767 m L AC after = 0.6403 − 0.2010 = 0.4393 m

− ! F W ! F AC θ AC ! F AB θ AB θ W We can also put the origin at A and use coordinates. ! C = − 0.6 0.5 ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ ! B = 0.4 0.5 ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ ! F W = 0 − mg ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = 0 − 392.4 ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ ! F AC + ! F AB + ! F W = 0 → ! C L AC F AC + ! B L AB F AB = − ! F W → ! C F AC L AC + ! B F AB L AB = − ! F W − 0.6 0.5 ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ F AC L AC + 0.4 0.5 ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ F AB L AB = 0 392.4 ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ → − 0.6 0.4 0.5 0.5 ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ F AC L AC F AB L AB ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = 0 392.4 ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ → − 0.6 0.4 0 0.5 0.5 392.4 ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ then find the reduced row-echelon form 1 0 313.92 0 1 470.88 ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ Multiply by the lengths: F AC = 313.92 ⋅ L AC = 245.17 F AB = 470.88 ⋅ L AB = 301.50 The rest is the same as above. This is a simple enough problem that we can also find the forces by geometry. θ AC = tan − 1 0.6 0.5 = 50.19 ° θ AB = tan − 1 0.4 0.5 = 38.66 ° θ W = 180 ° − θ AC − θ AB = 91.15 ° By Law of Sines, F W sin θ W = F AB sin θ AC = F AC sin θ AB F AB = F W sin θ AC sin θ W = 392.4 sin50.19 ° sin91.15 ° = 301.49 F AC = F W sin θ AB sin θ W = 392.4 sin38.66 ° sin91.15 ° = 245.18 The rest is the same as above.

p170-w3.1-sols

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