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The Force of Gravitational Attraction Between a Thin Rod of Mass

Question 16

Multiple Choice

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 GmMA(L+A) \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 MLdx\frac{M}{L} d x .Apply the formula above to each of the pieces, then form a Riemann sum.You know the rest…]  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) }


A) F=AA+LGM2dxLx(L+x) F=\int_{A}^{A+L} \frac{G M^{2} d x}{L x(L+x) }
B) F=AA+LGM2dxx(L+x) F=\int_{A}^{A+L} \frac{G M^{2} d x}{x(L+x) }
C) F=0AGM2dxL(L+x) F=\int_{0}^{A} \frac{G M^{2} d x}{L(L+x) }
D) F=0AGM2dxx(L+x) F=\int_{0}^{A} \frac{G M^{2} d x}{x(L+x) }

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