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Unformatted text preview: hanna (brh687) – hk 12 – Opyrchal – (11105) 1 This printout should have 14 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 3) 10.0 points A 1448.0 N uniform boom of length ℓ is sup ported by a cable, as shown. The boom is pivoted at the bottom, the cable is attached a distance 3 4 ℓ from the pivot, and a 3506.0 N weight hangs from the boom’s top. F T 3506 N 24 ◦ 66 ◦ Note: Figure is not drawn to scale a) Find the force F T applied by the sup porting cable. Correct answer: 2293 . 99 N. Explanation: Basic Concept: The second (rotational) condition of equi librium (axis of rotation at the base of the beam and beam length ℓ ): F T parenleftbigg 3 4 ℓ parenrightbigg F g,b parenleftbigg ℓ 2 parenrightbigg (cos θ b ) F g,m ℓ (cos θ b ) = 0 Given: F g,b = 1448 . 0 N F g,m = 3506 . 0 N θ b = 66 ◦ θ t = 24 ◦ Solution: 3 F T 2 F g,b (cos θ b ) 4 F g,m (cos θ b ) = 0 F T = (2 F g,b + 4 F g,m )(cos θ b ) 3 = [2(1448 N) + 4(3506 N)](cos 66 ◦ ) 3 = 2293 . 99 N . 002 (part 2 of 3) 10.0 points b) Find the horizontal component of the reac tion force on the bottom of the boom. Correct answer: 2095 . 67 N. Explanation: Basic Concept: The first (translational) condition of equi librium: Horizontally, F x = R x,base F T (cos θ t ) = 0 Solution: R x,base = F T (cos θ t ) = (2293 . 99 N)(cos24 ◦ ) = 2095 . 67 N . 003 (part 3 of 3) 10.0 points c) Find the vertical component of the reaction force on the bottom of the boom. Correct answer: 4020 . 95 N. Explanation: Basic Concept: The first (translational) condition of equi librium: Vertically, F y = R y,base + F T (sin θ t ) F g,m F g,b = 0 Solution: R y,base = F g,m + F g,b F T (sin θ t ) = 1448 N + 3506 N (2293 . 99 N)(sin 24 ◦ ) = 4020 . 95 N . 004 10.0 points The system shown in the figure is in equilib rium. A 16 kg mass is on the table. A string hanna (brh687) – hk 12 – Opyrchal – (11105) 2 attached to the knot and the ceiling makes an angle of 63 ◦ with the horizontal. The coeffi cient of the static friction between the 16 kg mass and the surface on which it rests is 0 . 36. The acceleration of gravity is 9 . 8 m / s 2 . 16 kg m 6 3 ◦ What is the largest mass m can have and still preserve the equilibrium? Correct answer: 11 . 3046 kg....
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 Spring '12
 James
 Physics, Force, Correct Answer, kg, Hanna

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