ssm_ch05

# ssm_ch05 - Chapter 5 Student Solutions Manual 5 We denote...

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

Chapter 5 – Student Solutions Manual 5. We denote the two forces G G F 1 and F 2 1 . According to Newton’s second law, GG G G G G FFm a Fm aF 12 2 += =s o ,. (a) In unit vector notation G F 1 20 0 = . ± Ni bg and () ( ) ( )( ) 22 2 ˆˆ ˆ 12.0 sin 30.0 m/s i 12.0 cos 30.0 m/s 6.00 m/s i 10.4m/s j. j a =− ° ° = − G 2 ˆ Therefore, ( ) ( ) 2 2.00kg 6.00 m/s i 2.00 kg 10.4 m/s j 20.0 N i 32.0 N i 20.8 N j. F + G ˆ (b) The magnitude of is G F 2 2 2 2 | | ( 32.0) ( 20.8) 38.2 N. xy FF F =+ = + = G (c) The angle that makes with the positive x axis is found from G F 2 tan θ = ( F 2 y / F 2 x ) = [(–20.8)/(–32.0)] = 0.656. Consequently, the angle is either 33.0° or 33.0° + 180° = 213°. Since both the x and y components are negative, the correct result is 213°. An alternative answer is . 213 360 147 °− °=− ° 13. (a) – (c) In all three cases the scale is not accelerating, which means that the two cords exert forces of equal magnitude on it. The scale reads the magnitude of either of these forces. In each case the tension force of the cord attached to the salami must be the same in magnitude as the weight of the salami because the salami is not accelerating. Thus the scale reading is mg , where m is the mass of the salami. Its value is (11.0 kg) (9.8 m/s 2 ) = 108 N. 19. (a) Since the acceleration of the block is zero, the components of the Newton’s second law equation yield T mg sin = 0 F N mg cos = 0. Solving the first equation for the tension in the string, we find Tm g == ° = sin . . 85 9 8 30 42 2 kg m / s N . ch

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

View Full Document
(b) We solve the second equation in part (a) for the normal force F N : ( )( ) 2 cos 8.5 kg 9.8 m/s cos 30 72 N . N Fm g θ == ° = (c) When the string is cut, it no longer exerts a force on the block and the block accelerates. The x component of the second law becomes – mg sin = ma , so the acceleration becomes 2 sin 9.8 sin 30 4.9 m/s . ag =− °= − The negative sign indicates the acceleration is down the plane. The magnitude of the acceleration is 4.9 m/s 2 . 25. In terms of magnitudes, Newton’s second law is F = ma , where F = G F net , || aa = G , and m is the (always positive) mass. The magnitude of the acceleration can be found using constant acceleration kinematics (Table 2-1). Solving v = v 0 + at for the case where it starts from rest, we have a = v / t (which we interpret in terms of magnitudes, making specification of coordinate directions unnecessary). The velocity is v = (1600 km/h) (1000 m/km)/(3600 s/h) = 444 m/s, so F 500 444 18 12 10 5 kg ms s N. bg . . × 29. The acceleration of the electron is vertical and for all practical purposes the only force acting on it is the electric force. The force of gravity is negligible. We take the + x axis to be in the direction of the initial velocity and the + y axis to be in the direction of the electrical force, and place the origin at the initial position of the electron. Since the force and acceleration are constant, we use the equations from Table 2-1: x = v 0 t and ya t F m t F H G I K J 1 2 1 2 22 .
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 04/09/2008 for the course PHYS 131 taught by Professor Drty during the Spring '08 term at Lafayette.

### Page1 / 10

ssm_ch05 - Chapter 5 Student Solutions Manual 5 We denote...

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

View Full Document
Ask a homework question - tutors are online