Traditional Another Old Midterm 2

Traditional Another Old Midterm 2 - Salamah(jms5723 –...

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Unformatted text preview: Salamah (jms5723) – Practice Midterm 02 – Yao – (58180) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Three blocks are on a frictionless horizontal surface. The bocks are connected by massless strings with tensions T ℓ and T r . 3 kg 5 kg 1 kg T ℓ T r 85 N 40 N Calculate the tension T ℓ . 1. T ℓ = 43 N 2. T ℓ = 8 N 3. T ℓ = 9 N 4. T ℓ = 40 N 5. T ℓ = 32 N 6. T ℓ = 69 N 7. T ℓ = 70 N correct 8. T ℓ = 72 N 9. T ℓ = 52 N 10. T ℓ = 47 N Explanation: Let m 1 = 3 kg m 2 = 5 kg m 3 = 1 kg F 1 = 85 N F 2 = 40 N Since the sum of the forces equals the total mass times the acceleration, we have summationdisplay F i = summationdisplay m i a F 1- F 2 = ( m 1 + m 2 + m 3 ) a. Thus a = F 1- F 2 m 1 + m 2 + m 3 = 85 N- 40 N 3 kg + 5 kg + 1 kg = 5 m / s 2 . Isolating each section, we have T ℓ = F 1- m 1 a = 85 N- (3 kg) (5 m / s 2 ) = 70 N , and T r = F 1- ( m 1 + m 2 ) a = 85 N- (3 kg + 5 kg) (5 m / s 2 ) = 45 N . There are two versions of this problem ask- ing for either T ℓ and T r . 002 10.0 points Given: The friction between the block with mass 12 kg and the wedge with mass 16 kg is 0 . 41 . The surface between the wedge with mass 16 kg and the horizontal plane is smooth (without friction). The acceleration of gravity is 9 . 8 m / s 2 . A block is released on the inclined plane (top side of the wedge). 1 2 k g μ = . 4 1 36 ◦ 16 kg μ = 0 F 9 . 8m / s 2 What is the maximum force F which can be exerted on the 16 kg block in order that the 12 kg block does not move up the plane? Correct answer: 444 . 181 N. Salamah (jms5723) – Practice Midterm 02 – Yao – (58180) 2 Explanation: Let : m A = 16 kg , block A m B = 12 kg , block B θ = 36 ◦ , and μ s = 0 . 41 between A and B . Using Newton’s second law, the accelera- tion of the wedge of mass 16 kg and block of mass 12 kg is a AB = F m A + m B (1) Consider the free body diagram for block B m g s i n θ N μ N mg a AB The condition that block B does not move with respect to wedge A implies that its ac- celeration down the wedge with respect to the wedge is zero. Applying Newton’s second law for B in the direction parallel and perpendic- ular to the wedge yields summationdisplay F ⊥ : N - m B g cos θ = m B a AB sin θ (1) summationdisplay F bardbl : m B g sin θ + μ N = m B a AB cos θ (2) Using Eq. 1 to solve for the normal force N exerted on block B by block A , we have N = m B ( g cos θ + a AB sin θ ) . Substituting N from Eq. 1 into Eq. 2, we have m B g sin θ + μm B g cos θ + μm B a AB sin θ = m B a AB cos θ Solving for a AB , we have a AB = g sin θ + μ cos θ cos θ- μ sin θ = (9 . 8 m / s 2 ) sin 36 ◦ + 0 . 41cos 36 ◦ cos 36 ◦- . 41sin 36 ◦ = (9 . 8 m / s 2 ) . 587785 + 0 . 331697 . 809017- . 240992 = 15 . 8636 m / s 2 ....
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Traditional Another Old Midterm 2 - Salamah(jms5723 –...

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