Question

calculate the resultant of the following forces at a point, making use of resolution process. (i) 100 root2 dyne along north east (ii) 980 root2 dyne along north west (iii) 1960 dyne along south please show working ...

Answer 1

we have to calculate the resultant of the following forces at a point, making use of resolution process.

(i) 100√2 Dyne along north east

(ii) 980√2 Dyne along north west

(iii) 1960 Dyne along south

solution : write all forces in vector forms,

(i) 100√2 Dyne along north east,

$F_1=100\sqrt{2}(cos45^{\circ}\hat i+sin45^{\circ}\hat j)$

⇒$F_1=100\hat i+100\hat j$

(ii) 980√2 Dyne along north west

$F_2=980\sqrt{2}(-cos45^{\circ}\hat i+sin45^{\circ}\hat j)$

⇒$F_2=-980\hat i+980\hat j$

(iii) 1960 Dyne along south

$F_3=-1960\hat j$

now adding the all forces and getting resultant force.

$F=F_1+F_2+F_3$

⇒$F=(100\hat i+100\hat j)+(-980\hat i+980\hat j)+(-1960)\hat j$

⇒$F=-880\hat i-880\hat j$

Therefore the magnitude of resultant force is √{880² + 880²} = 880√2 Dyne directed along south west.

The resultant of forces is shown in figure.