phys1hw5 - homework 05 – ANDERSON, ZACH – Due: Sep 29...

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

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: homework 05 – ANDERSON, ZACH – Due: Sep 29 2007, 4:00 am 1 Question 1, chap 5, sect 6. part 1 of 1 10 points A block of mass m accelerates with acceler- ation g up a frictionless plane that is inclined at an angle α above the horizontal. The force F o that pushes the block is at an angle β above the horizontal. Find the force F o . m α β F o 1. mg 1 − sin β cos( β − α ) 2. mg 1 cos( α + β ) 3. mg 1 + sin α cos( α + β ) correct 4. mg 1 + sin α cos( β − α ) 5. mg 1 + sin α cos( α − β ) 6. mg 1 − sin β cos( α + β ) 7. mg sin β cos( α + β ) 8. mg 1 + sin β cos( β − α ) 9. mg 1 + sin β cos( α + β ) 10. mg 1 + sin β cos( α − β ) Explanation: The acceleration up the plane is given, a = g . The angle that F o makes with respect to the inclined plane is α + β . So the component of F o which is parallel to the inclined plane is F o cos( α + β ) Therefore F o cos( α + β ) − mg sin α = ma = mg F o = mg 1 + sin α cos( α + β ) Question 2, chap 6, sect 1. part 1 of 3 10 points The suspended 2 . 7 kg mass on the right is moving up, the 2 . 3 kg mass slides down the ramp, and the suspended 8 . 5 kg mass on the left is moving down. The coefficient of friction between the block and the ramp is 0 . 12 . The acceleration of gravity is 9 . 8 m / s 2 . The pulleys are massless and frictionless. 2 . 3 k g μ = . 1 2 32 ◦ 8 . 5 kg 2 . 7 kg What is the acceleration of the three block system? Correct answer: 4 . 92523 m / s 2 (tolerance ± 1 %). Explanation: Let : m 1 = 2 . 7 kg , m 2 = 2 . 3 kg , m 3 = 8 . 5 kg , and θ = 32 ◦ . Basic Concept: F net = ma negationslash = 0 homework 05 – ANDERSON, ZACH – Due: Sep 29 2007, 4:00 am 2 Solution: The acceleration a of each mass is the same, but the tensions in the two strings will be different. Let T 1 be the tension in the right string and T 3 the tension in the left string. Consider the free body diagrams for each mass T 3 m 3 g a T 1 m 1 g a T 3 T 1 N μ N a m 2 g For the mass m 1 , T 1 acts up and the weight m 1 g acts down, with the acceleration a di- rected upward F net 1 = m 1 a = T 1 − m 1 g (1) For the mass on the table, the parallel compo- nent of its weight is mg sin θ and the perpen- dicular component of its weight is mg cos θ . ( N = mg cos θ from equilibrium). The accel- eration a is directed down the table, T 3 and the parallel weight component m 2 g sin θ act down the table, and T 1 and the frictional force μ N = μm 2 g cos θ act up the table F net 2 = m 2 a (2) = T 3 + m 2 g sin θ − T 1 − μm 2 g cos θ . For the mass m 3 , T 3 acts up and the weight m 3 g acts down, with the acceleration a di- rected downward F net 3 = m 3 a = m 3 g − T 3 . (3) Adding Eqs. (1), (2), & (3) yields ( m 1 + m 2 + m 3 ) a = m 3 g + m 2 g sin θ − μm 2 g cos θ − m 1 g ....
View Full Document

This note was uploaded on 04/10/2008 for the course PHY 303K taught by Professor Turner during the Fall '08 term at University of Texas.

Page1 / 13

phys1hw5 - homework 05 – ANDERSON, ZACH – Due: Sep 29...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online