Question

(A)   Draw the FBD of block A

(B)   Draw  the FBD of block B

(C)   What is the weight of the hanging block that will maintain static equilibrium?

Answer 1

A D B इ०० 4 W W E-0 → T= N/Cer 70+ AN/sin 10° . (A) FBD Of A > (१) yिo > My) + AN, Cek10 = N, Sind - N) = ||2.5N, From O →ा- 59.62N X म 20 >AN 7N, 70° 10 m१) JAN, O =» From-2 →

FBD of Block B :- B) juni Ni -Tz M. Na mBeg) NA Ni 2 Tz MNIK 60° 700 20° 30 38 ظ ki Yng MN2 Efn=0 T2CO3 30°= N2 COS 60 +14 , N, 40520 +M Cos30° +N, Costo Efyao y maq) +M, N,Sing0 + N, Sint0= N2 Sin 60" + M, N, Sin 20+ Tzsin30° 2019,81) + 0.1(N2)+ 112.551970 = N2 (2)+0.28112.5 Sin28+ Tz 196.2 + 0.05N2 + 105.72 = 0.87N27 7.69 + T2 294,23 - 0.87N2 -0,05 N2 +0.552 0.82 N2 +0.5 T2 = 294.23 - 0

0.82 N2 to 5 T2 +294.23=0 T() = +20.2) (12:45) 00520 +6.1X asr ) + 125676 0.87T2 = 0.586N2 + 21,14 + 38.48 0.87T2 = 0.586 N₂ + 59,62 0.586 N2 -0.87T2+59.62=0 N2= 219, 04 N T2 = 229,24 N m equilibrium;- c) system to be Ti R R 600 W R = 7,² + 73 7 27; T3 cos 60" Ti + MN = T3 N=8112:57 7 73 27 112:573 T3 = 146.20 N → T3= mg m= 14.90kg .

Answer 2

A D B इ०० 4 W W E-0 → T= N/Cer 70+ AN/sin 10° . (A) FBD Of A > (१) yिo > My) + AN, Cek10 = N, Sind - N) = ||2.5N, From O →ा- 59.62N X म 20 >AN 7N, 70° 10 m१) JAN, O =» From-2 →

FBD of Block B :- B) juni Ni -Tz M. Na mBeg) NA Ni 2 Tz MNIK 60° 700 20° 30 38 ظ ki Yng MN2 Efn=0 T2CO3 30°= N2 COS 60 +14 , N, 40520 +M Cos30° +N, Costo Efyao y maq) +M, N,Sing0 + N, Sint0= N2 Sin 60" + M, N, Sin 20+ Tzsin30° 2019,81) + 0.1(N2)+ 112.551970 = N2 (2)+0.28112.5 Sin28+ Tz 196.2 + 0.05N2 + 105.72 = 0.87N27 7.69 + T2 294,23 - 0.87N2 -0,05 N2 +0.552 0.82 N2 +0.5 T2 = 294.23 - 0

0.82 N2 to 5 T2 +294.23=0 T() = +20.2) (12:45) 00520 +6.1X asr ) + 125676 0.87T2 = 0.586N2 + 21,14 + 38.48 0.87T2 = 0.586 N₂ + 59,62 0.586 N2 -0.87T2+59.62=0 N2= 219, 04 N T2 = 229,24 N m equilibrium;- c) system to be Ti R R 600 W R = 7,² + 73 7 27; T3 cos 60" Ti + MN = T3 N=8112:57 7 73 27 112:573 T3 = 146.20 N → T3= mg m= 14.90kg .