The center sphere (2 kg) experiences gravitational forces from the other two spheres (10 kg and 5 kg). The force due to the 10 kg sphere acts to the left (negative x-direction), and the force due to the 5 kg sphere acts to the right (positive x-direction). The center sphere also experiences a gravitational force upwards (positive y-direction) and a gravitational force downwards (negative y-direction). Here the picture doesn't specify any other masses, so we only consider the two masses given in the horizontal plane.
The magnitude of the gravitational force between two objects is given by F=Gm1m2r2F = G\frac{m_1 m_2}{r^2}F=Gr2m1m2, where G is the gravitational constant, m1m_1m1 and m2m_2m2 are the masses of the objects, and rrr is the distance between their centers.
Force due to the 10 kg sphere: F10=G(2)(10)502=20G2500=2G250F_{10} = G\frac{(2)(10)}{50^2} = \frac{20G}{2500} = \frac{2G}{250}F10=G502(2)(10)=250020G=2502G Force due to the 5 kg sphere: F5=G(2)(5)202=10G400=G40F_5 = G\frac{(2)(5)}{20^2} = \frac{10G}{400} = \frac{G}{40}F5=G202(2)(5)=40010G=40G
Since G40>2G250\frac{G}{40} > \frac{2G}{250}40G>2502G, the force due to the 5 kg sphere is larger than the force due to the 10 kg sphere.
The free body diagram for the center sphere (2 kg) will have two arrows:
An arrow pointing to the left representing the force due to the 10 kg sphere (F10F_{10}F10). This arrow should be shorter.
An arrow pointing to the right representing the force due to the 5 kg sphere (F5F_5F5). This arrow should be longer.
<----F₁₀ 2 kg F₅----> 10 kg-------------------●-------------------5 kg 50 m 20 m
The free body diagram shows two horizontal forces acting on the 2 kg sphere. F5F_5F5 points towards the 5 kg mass and is longer than F10F_{10}F10 pointing to the 10 kg sphere.
<----F₁₀ 2 kg F₅---->
10 kg-------------------●-------------------5 kg 50 m 20 m
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