ch05-p019 - ) cos 30 = 72 N. (c) When the string is cut it...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
19 (a) The free-body diagram is shown in Fig. 5 16 of the text. Since the acceleration of the block is zero, the components of the Newton’s second law equation yield T mg sin θ =0and F N mg cos θ = 0. Solve the first equation for the tension force of the string: T = mg sin θ = (8 . 5 kg)(9 . 8m / s 2 )sin30 =42N . (b) Solve the second equation for F N : F N = mg cos θ =(8 . 5 kg)(9 . 8m / s
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ) cos 30 = 72 N. (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 a = g sin = (9 . 8 m / s 2 ) sin 30 = 4 . 9 m / s 2 . The negative sign indicates the acceleration is down the plane....
View Full Document

Ask a homework question - tutors are online