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Three forces are applied to a square plate as shown in Figure. Find the modulus, direction, and the point of application of the resultant force, if this point is taken on the side BC. |
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Answer» For coplanar forces, about any point in the same plane, `sumvecr_ixxvecF_i=vecrxxvecF_(n et)` (where `vecF_(n et)=sum vecF_i` = resultant force) or, `vecN_(n et)=vecrxxvecF_(n et)` Thus length of the arm, `l=(N_(n et))/(F_(n et))` Here obviously `|vecF_(n et)|=2F` and it is directed toward right along AC. Take the origin at C. Then about C, `vecN=(sqrt2aF+a/sqrt2F-sqrt2aF)` directed normally into the plane of figure. Here `a=` side of the square.) Thus `vecN=F(a)/(sqrt2)` directed into the plane of the figure. Hence `l=(F(a//sqrt2))/(2F)=(a)/(2sqrt2)=a/2sin45^@` Thus the point of application of force is at the mid point of the side BC. |
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