Calculate the resultant of the following forces acting at a point- making us of resolution process-i 100-2 newton along north-eastii 980-2 newton along north-westiii 1960 newton along south

F_r = #10;√(( 100√(2))^2 + ( 980√(2))^2) #10; #10; #10; = √( )#10;( 100^2× 2) + 980^2× 2 #10; #10;=1393 N #10; #10;Total = #10;19 ...

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Standard XII

Physics

Time Period of Pendulum

Calculate the...

Question

Calculate the resultant of the following forces acting at a point, making us of resolution process.
(i) $100 \sqrt{2}$ newton along north-east
(ii) $980 \sqrt{2}$ newton along north-west
(iii) 1960 newton along south

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Solution

$F_{r} = \sqrt{{(100 \sqrt{2})}^{2} + {(980 \sqrt{2})}^{2}}$

$= \sqrt{(100^{2} \times 2) + 980^{2} \times 2}$

=1393 N

Total = 1960 – 1390 ( South $F_{r}$ )

= 570

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Calculate the resultant of the following forces acting at a point, making us of resolution process.(i) 100√2 newton along north-east(ii) 980√2 newton along north-west(iii) 1960 newton along south

Fr=√(100√2)2+(980√2)2 =√(1002×2)+9802×2 =1393 N Total = 1960 – 1390 ( South Fr) = 570

Calculate the resultant of the following forces acting at a point, making us of resolution process.
(i) $100 \sqrt{2}$ newton along north-east
(ii) $980 \sqrt{2}$ newton along north-west
(iii) 1960 newton along south

$F_{r} = \sqrt{{(100 \sqrt{2})}^{2} + {(980 \sqrt{2})}^{2}}$

$= \sqrt{(100^{2} \times 2) + 980^{2} \times 2}$

=1393 N

Total = 1960 – 1390 ( South $F_{r}$ )

= 570