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statics 3

# statics 3 - Problem 3.1 The gure shows the external forces...

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Problem 3.1 The figure shows the external forces act- ing on an object in equilibrium. The forces F 1 = 32 N and F 3 = 50 N. Determine F 2 and the angle α . y x 30 ° 12 ° α F 1 F 3 F 2 Solution: Write the forces in component form. F 1 = 32 sin 30 i + 32 cos 30 j F 1 = 16 i + 27 . 7 j N F 2 = 50 cos 12 i 50 sin 12 j F 2 = 48 . 9 i 10 . 4 j (N) F 2 = F 2 cos α i F 2 sin α j Sum components in x and y directions F x = 16 48 . 9 + F 2 cos α = 0 F y = 27 . 7 10 . 4 F 2 sin α = 0 Solving, we get F z = 37 . 2 N α = 27 . 73 y x F 1 F 2 F 3 30 ° 12 ° α Problem 3.2 The force F 1 = 100 N and the angle α = 60 . The weight of the ring is negligible. Determine the forces F 2 and F 3 . x y 30 ° α F 3 F 2 F 1 Solution: Write the forces in component form. F 1 = F 1 i + 0 j F 2 = F 2 cos 30 i + F 2 sin 30 j F 3 = F 3 cos α i F 3 sin α j We know F = 0 , thus F x = 0 and F y = 0 . Writing the equilibrium equations, we have F x = F 1 F 2 cos 30 F 3 cos α = 0 F y = F 2 sin 30 F 3 sin α = 0 F 1 = 100 N , α = 60 Solving, we get F 2 = 86 . 6 N , F 3 = 50 N x y 30 ° α F 3 F 2 F 1

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Problem 3.3 Consider the forces shown in Prob- lem 3.2. Suppose that F 2 = 100 N and you want to choose the angle α so that the magnitude of F 3 is a min- imum. What is the resulting magnitude of F 3 ? Strategy: Draw a vector diagram of the sum of the three forces. Solution: | F 2 | = 100 N, F 1 is horizontal, and F = 0 . From the diagram, α = 90 and | F 3 | = 50 N F 1 F 3 min F 2 α 60 ° 30 ° x Problem 3.4 The beam is in equilibrium. If A x = 77 kN, B = 400 kN, and the beam’s weight is negligible, what are the forces A y and C ? A y B 30 ° C A x 2 m 4 m Solution: + F x = A x C sin 30 = 0 + F y = A y B + C cos 30 = 0 A x = 77 kN , B = 400 kN Solving, we get A y = 267 kN C = 154 kN A y B 30 ° C A x 2 m 4 m Problem 3.5 Suppose that the mass of the beam shown in Problem 3.4 is 20 kg and it is in equilibrium. The force A y points upward. If A y = 258 kN and B = 240 kN, what are the forces A x and C ? Solution: + F x = 0 = A x C sin 30 = 0 + F y = 0 = A y B (20)(9 . 81) + C cos 30 = 0 A y = 258 kN , B = 240 kN Solving, we get A x = 103 kN C = 206 kN A x A y B C 30 ° (20 kg) (9.81)m/s 2
Problem 3.6 A zoologist estimates that the jaw of a predator, Martes , is subjected to a force P as large as 800 N. What forces T and M must be exerted by the temporalis and masseter muscles to support this value of P ? T 22 ° P M 36 ° Solution: Resolve the forces into scalar components, and solve the equilibrium equations . . . Express the forces in terms of horizontal and vertical unit vectors: T = | T | ( i cos 22 + j sin 22 ) = | T | (0 . 927 i + 0 . 375 j ) P = 800( i cos 270 + j sin 270 ) = 0 i 800 j M = | M | ( i cos 144 + j sin 144 ) = | M | ( 0 . 809 i + 0 . 588 j ) Apply the equilibrium conditions, F = 0 = T + M + P = 0 Collect like terms: F x = (0 . 927 | T | − 0 . 809 | M | ) i = 0 F y = (0 . 375 | T | − 0 . 588 | M | − 800) j = 0 Solve the first equation, | T | = ( 0 . 809 0 . 927 ) | M | = 0 . 873 | M | Substitute this value into the second equation, reduce algebraically, and solve: | M | = 874 N , | T | = 763 . 3 N T 22 ° P M 36 °

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Problem 3.7 The two springs are identical, with un- stretched lengths 250 mm and spring constants k = 1200 N/m.
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statics 3 - Problem 3.1 The gure shows the external forces...

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