[amp_mcq option1=”5.5 kg” option2=”6.25 kg” option3=”8.75 kg” option4=”10.5 kg” correct=”option4″]<\/p>\n
The correct answer is $\\boxed{\\text{B}}$.<\/p>\n
Let $a$ be the distance of the resultant from the 20 kg force and $b$ be the distance of the resultant from the 15 kg force. We know that the resultant of the two forces is 5 kg, so the equation $20a+15b=5$ holds. We also know that the distance of the resultant from the 20 kg force is to be the same as the distance of the former resultant was from the 15 kg force, so $a=b$. Substituting this into the equation $20a+15a=5$, we get $35a=5$. Solving for $a$, we get $a=\\frac{5}{35}=\\frac{1}{7}$.<\/p>\n
Now, we know that the 20 kg force is diminished by $x$ kg, so the new force is 20-x kg. The equation $20(a)+(15-x)b=5$ now holds. Substituting $a=\\frac{1}{7}$ into this equation, we get $20\\left(\\frac{1}{7}\\right)+(15-x)b=5$. Solving for $b$, we get $b=\\frac{105-20x}{105}$.<\/p>\n
We want the distance of the resultant from the 20 kg force to be the same as the distance of the former resultant was from the 15 kg force, so we want $a=b$. Substituting $b=\\frac{105-20x}{105}$ into the equation $a=b$, we get $\\frac{1}{7}=\\frac{105-20x}{105}$. Solving for $x$, we get $x=\\boxed{6.25}$.<\/p>\n","protected":false},"excerpt":{"rendered":"
[amp_mcq option1=”5.5 kg” option2=”6.25 kg” option3=”8.75 kg” option4=”10.5 kg” correct=”option4″]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[645],"tags":[],"class_list":["post-6686","post","type-post","status-publish","format-standard","hentry","category-applied-mechanics-and-graphic-statics","no-featured-image-padding"],"yoast_head":"\n