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A force of 10.5 N acts on a particle along a direction making an angle of `37^0` with the vertical. Find the component of the force in the vertical direction. |
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Answer» The component of the force in the vertical directiionwill be `F_ _|_= F cos theta =(10.5N)(cos37^0)` `=(10.5N) 4/5 =8.40N` We can easily add two or more vectors if we know their components along the rectangular cordiN/Ate xes. Let us have ` veca=a_x veci+a_yvecj+a_2veck` `vecb=b_x veci+b_yvecj+b_zveck` `vecc=c_xveci+c_yvecj+c_2veck` then `veca+vecb+vecc=(a_x+b_x+c_x)veci+(a_y+by+c_y)vecj+(a_z+b_z+c_z)veck`. If all the vectors are in the X-Y plane then all the z components are zero and the resultant is simply `veca+vecb+vecc=(a_x+b_x+c_x) veci+(a_y+b_y+c_y)vecj` This is the sum of twomutually perpendicular vectors of magnitude `(a_x+b_x+c_x) and (a_y+b_y+c_y)`. The resutant can easily be found to hae a magnitude `sqrt((a_x+b_x+c_x)^2+(a_y+b_y+c_y)^2)` making an angle alpha with the X-axis whre `tan alpha= (a_y+b_y+c_y)/(a_x+b_x+c_x)` |
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