The force of gravitational attraction between a thin rod of mass M and length L and a particle of mass m lying on the same line as the rod at a distance of A from one of the ends is $\frac{G m M}{A(L+A)}$ .Use this result to select an integral for the total force due to gravity between two thin rods, both of mass M and length L, lying along the same line and separated by a distance A.[Hint: Divide one of the rods into small pieces, each of length dx and mass $\frac{M}{L} d x$ .Apply the formula above to each of the pieces, then form a Riemann sum.You know the rest…]
A) $F=\int_{A}^{A+L} \frac{G M^{2} d x}{L x(L+x)}$
B) $F=\int_{A}^{A+L} \frac{G M^{2} d x}{x(L+x)}$
C) $F=\int_{0}^{A} \frac{G M^{2} d x}{L(L+x)}$
D) $F=\int_{0}^{A} \frac{G M^{2} d x}{x(L+x)}$

Question

The force of gravitational attraction between a thin rod of mass M and length L and a particle of mass m lying on the same line as the rod at a distance of A from one of the ends is $\frac{G m M}{A(L+A)}$ .Use this result to select an integral for the total force due to gravity between two thin rods, both of mass M and length L, lying along the same line and separated by a distance A.[Hint: Divide one of the rods into small pieces, each of length dx and mass $\frac{M}{L} d x$ .Apply the formula above to each of the pieces, then form a Riemann sum.You know the rest…] 
A) $F=\int_{A}^{A+L} \frac{G M^{2} d x}{L x(L+x)}$ 
B) $F=\int_{A}^{A+L} \frac{G M^{2} d x}{x(L+x)}$ 
C) $F=\int_{0}^{A} \frac{G M^{2} d x}{L(L+x)}$ 
D) $F=\int_{0}^{A} \frac{G M^{2} d x}{x(L+x)}$

Quizplus · Accepted Answer

The Answer of The force of gravitational attraction between a thin rod of mass M and length L and a particle of mass m lying on the same line as the rod at a distance of A from one of the ends is $\frac{G m M}{A(L+A)}$ .Use this result to select an integral for the total force due to gravity between two thin rods, both of mass M and length L, lying along the same line and separated by a distance A.[Hint: Divide one of the rods into small pieces, each of length dx and mass $\frac{M}{L} d x$ .Apply the formula above to each of the pieces, then form a Riemann sum.You know the rest…] 
A) $F=\int_{A}^{A+L} \frac{G M^{2} d x}{L x(L+x)}$ 
B) $F=\int_{A}^{A+L} \frac{G M^{2} d x}{x(L+x)}$ 
C) $F=\int_{0}^{A} \frac{G M^{2} d x}{L(L+x)}$ 
D) $F=\int_{0}^{A} \frac{G M^{2} d x}{x(L+x)}$

The Force of Gravitational Attraction Between a Thin Rod of Mass