# Hw 5 - A 200-N force is applied as shown to the bracket ABC...

This preview shows pages 1–4. Sign up to view the full content.

PROBLEM 3.1 A foot valve for a pneumatic system is hinged at B. Knowing that a = 28°, detennine the moment of the 16-N force about Point B by resolving the force into horizontal and vertical components. SOLUTION Note that f} =a-2Do = 28"_20° =8" and F~ = (16 N)cos8 D = 15.&443 N F x = (16 N)sin 8" = 2.2268 N Also x = (0.17 m)cos20° = 0.159748 m y = (0.17 m)sin20° = 0.058143 m Noting that the direction of the moment of eaeh force component about B is counterclockwise. M B =xF y + yF x =(0.159748 rnX2.2268 N) +(0.058143 rn)(I5.8443 N) =1.277N·m or M B =l.277N.m) . ....

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

View Full Document
PROBLEM 3.2 A foot valve for II pneumatie system is hinged at B. Knowing that a = 28<l, detennine the moment of the 16-N force about Point B by resolving the force into components along ABC and in a direction perpendicular to ABC. SOLUTION First resolve the 4-lb force into components P and Q. where Q=(16 N)sin2BO =7.5115N Then M B = rA1BQ = (0.17 m)(7.5115 N) =1.277N-m
PROBLEM 3.21

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: A 200-N force is applied as shown to the bracket ABC. Detennine the moment ofthe force about A. SOLUTION We have where ["CIA = (0.06 m)1 + (0.075 m)j Fe = -(200 N)cos 30"j + (200 N)sin 30"k Then I M A ~ 200 0.06 o j 0.Q75 -cos30" k 0 sin 30" ~ 200[(0.075,io 30°)1 -(0.06';0 30 0 )j - (0.06co, 30 0 )k] or M. ~(7.50N·m)i-(6.00N·m)j-(i039N·m)k <II PROBLEM 3.26 A small boat hangs from two davits, one of whieh is shown in the figure. The tension in line ABAD is 82 lb. Determine the moment about C of the resultant foree R A exerted on the davit at A. SOLUTION We have where and Thus Also U,ing Eq. (3.21), R A =2F,(8+F AD F,a =-(821b)j F =F AD=(82Ib)61-7.75 1-3k AD AD AD 10.25 F'D = (48 Ib)i - (62 Ib)J -(24Ib)k R, = 2F,. + F'D = (48 Ib)l- (226 Ib)J- (24Ib)k r"e =(7.75 ft)j+(3 ft)k i j k Me = 0 7.75 3 48 -226 -24 = (492 lb· ft)1 + (1441b· ft)j - (372 lb· ft)k Me =(492 lb· ft)1 + (144.0 Ib·ft)J-(372lb·ft)k . .....
View Full Document

## This note was uploaded on 10/05/2009 for the course CE statics taught by Professor Epstein during the Spring '09 term at UConn.

### Page1 / 4

Hw 5 - A 200-N force is applied as shown to the bracket ABC...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online