{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

solution11_pdf

# solution11_pdf - arellano(aa39398 Homework 7 Weathers(17101...

This preview shows pages 1–3. Sign up to view the full content.

arellano (aa39398) – Homework 7 – Weathers – (17101) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points In the figure below the left-hand cable has a tension T 1 and makes an angle of 40 with the horizontal. The right-hand cable has a tension T 3 and makes an angle of 46 with the horizontal. The horizontal cable connecting the two masses has a tension of 79 N. M 1 M 2 79 N T 1 T 3 46 40 Determine the mass M 2 . The acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 8 . 35 kg. Explanation: Let : W 1 = M 1 g , W 2 = M 2 g , θ 1 = 40 , θ 3 = 46 , and T 2 = 79 N . T 3 T 1 θ 3 θ 1 T 3 cos θ 3 T 1 cos θ 1 W 2 W 1 Note: T 1 cos θ 1 = T 2 = T 3 cos θ 3 Basic Concepts: summationdisplay vector F = mvectora = 0 vector W = mvectorg Solution: Consider the point of attachment of cable 2 and cable 3. Vertically, W 2 = M 2 g acts down and T 3 sin θ 3 acts up, so F net = W 2 T 3 sin θ 3 = 0 = T 3 sin θ 3 = W 2 . (1) Horizontally, T 2 acts to the left and T 3 cos θ 3 acts to the right, so F net = T 2 T 3 cos θ 3 = 0 = T 3 cos θ 3 = T 2 . (2) Dividing Eq. 1 by Eq. 2, we have tan θ 3 = W 2 T 2 . W 2 = T 2 tan θ 3 = (79 N) tan46 = 81 . 8069 N M 2 = W 2 g = 81 . 8069 N 9 . 8 m / s 2 = 8 . 35 kg and by symmetry , we have T 1 cos θ 1 T 3 cos θ 3 = 0 , so W 1 = T 2 tan θ 1 = (79 N) tan40 = 103 . 127 N M 1 = W 1 g = 66 . 2889 N 9 . 8 m / s 2 = 6 . 76 kg . 002 (part 1 of 2) 10.0 points A pulley is massless and frictionless. 2 kg, 2 kg, and 8 kg masses are suspended as in the figure. 3 . 2 m 24 . 6 cm ω 2 kg 2 kg 8 kg T 2 T 1 T 3

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
arellano (aa39398) – Homework 7 – Weathers – (17101) 2 What is the tension T 1 in the string be- tween the two blocks on the left-hand side of the pulley? The acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 26 . 1333 N. Explanation: Let : R = 24 . 6 cm , m 1 = 2 kg , m 2 = 2 kg , m 3 = 8 kg , and h = 3 . 2 m . Consider the free body diagrams 2 kg 2 kg 8 kg T 1 T 2 T 3 m 1 g T 1 m 2 g m 3 g a a For each mass in the system vector F net = mvectora . Since the string changes direction around the pulley, the forces due to the tensions T 2 and T 3 are in the same direction (up). The acceleration of the system will be down to the right ( m 3 > m 1 + m 2 ), and each mass in the system accelerates at the same rate (the string does not stretch). Let this acceleration rate be a and the tension over the pulley be T T 2 = T 3 . For the lower left-hand mass m 1 the accel- eration is up and T 1 m 1 g = m 1 a . (1) For the upper left-hand mass m 2 the acceler- ation is up and T T 1 m 2 g = m 2 a . (2) For the right-hand mass m 3 the acceleration is down and T + m 3 g = m 3 a . (3) Adding Eqs. (1), (2), and (3), we have ( m 3 m 1 m 2 ) g = ( m 1 + m 2 + m 3 ) a (4) a = m 3 m 1 m 2 m 1 + m 2 + m 3 g (5) = 8 kg 2 kg 2 kg 2 kg + 2 kg + 8 kg (9 . 8 m / s 2 ) = 3 . 26667 m / s 2 . The tension in the string between block m 1 and block m 2 (on the left-hand side of the pulley) can be determined from Eq. (1).
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern