Reaction forces on the top sphere

Classical Mechanics Level pending

Three spheres of radii 5 , 6 , 7 5, 6, 7 are placed on a horizontal table, tangent to each other. They are then fixed in their position (for example, by using glue or welding). Now a fourth sphere of radius 8 8 is placed on top of the three spheres. If this top sphere weighs 100 100 Newtons, find the sum of the magnitudes of the reaction forces from the three bottom spheres, acting on the top sphere and balancing its weight force.


The answer is 109.12.

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1 solution

Steven Chase
Feb 21, 2021

Begin by writing down the known and unknown quantities. The coordinates are for the centers of the spheres. The table is considered to be the x y xy plane.

R 5 = 5 R 6 = 6 R 7 = 7 R 8 = 8 ( x 5 , y 5 , z 5 ) = ( 0 , 0 , R 5 ) ( x 6 , y 6 , z 6 ) = ( ? , 0 , R 6 ) ( x 7 , y 7 , z 7 ) = ( ? , ? , R 7 ) ( x 8 , y 8 , z 8 ) = ( ? , ? , ? ) R_5 = 5 \\ R_6 = 6 \\ R_7 = 7 \\ R_8 = 8 \\ (x_5, y_5, z_5) = (0,0,R_5) \\ (x_6, y_6, z_6) = (? ,0,R_6) \\ (x_7, y_7, z_7) = (? , ?,R_7) \\ (x_8, y_8, z_8) = (? ,?,?)

Solve the following six nonlinear equations for ( x 6 , x 7 , y 7 , x 8 , y 8 , z 8 ) (x_6, x_7, y_7, x_8, y_8, z_8) .

( x 5 x 6 ) 2 + ( y 5 y 6 ) 2 + ( z 5 z 6 ) 2 = ( R 5 + R 6 ) 2 ( x 5 x 7 ) 2 + ( y 5 y 7 ) 2 + ( z 5 z 7 ) 2 = ( R 5 + R 7 ) 2 ( x 5 x 8 ) 2 + ( y 5 y 8 ) 2 + ( z 5 z 8 ) 2 = ( R 5 + R 8 ) 2 ( x 6 x 7 ) 2 + ( y 6 y 7 ) 2 + ( z 6 z 7 ) 2 = ( R 6 + R 7 ) 2 ( x 6 x 8 ) 2 + ( y 6 y 8 ) 2 + ( z 6 z 8 ) 2 = ( R 6 + R 8 ) 2 ( x 7 x 8 ) 2 + ( y 7 y 8 ) 2 + ( z 7 z 8 ) 2 = ( R 7 + R 8 ) 2 (x_5 - x_6)^2 + (y_5 - y_6)^2 + (z_5 - z_6)^2 = (R_5 + R_6)^2 \\ (x_5 - x_7)^2 + (y_5 - y_7)^2 + (z_5 - z_7)^2 = (R_5 + R_7)^2 \\ (x_5 - x_8)^2 + (y_5 - y_8)^2 + (z_5 - z_8)^2 = (R_5 + R_8)^2 \\ (x_6 - x_7)^2 + (y_6 - y_7)^2 + (z_6 - z_7)^2 = (R_6 + R_7)^2 \\ (x_6 - x_8)^2 + (y_6 - y_8)^2 + (z_6 - z_8)^2 = (R_6 + R_8)^2 \\ (x_7 - x_8)^2 + (y_7 - y_8)^2 + (z_7 - z_8)^2 = (R_7 + R_8)^2

Having found the centers of all spheres, calculate unit vectors u 58 \vec{u}_{5 8} , u 68 \vec{u}_{6 8} , and u 78 \vec{u}_{7 8} from the centers of spheres 5 5 , 6 6 , and 7 7 to the center of sphere 8 8 . These are the lines of action for the forces. Then solve the following linear equations for the forces:

F 58 u 58 x + F 68 u 68 x + F 78 u 78 x = 0 F 58 u 58 y + F 68 u 68 y + F 78 u 78 y = 0 F 58 u 58 z + F 68 u 68 z + F 78 u 78 z = 100 F_{58} u_{58x} + F_{68} u_{68x} + F_{78} u_{78x} = 0 \\ F_{58} u_{58y} + F_{68} u_{68y} + F_{78} u_{78y} = 0 \\ F_{58} u_{58z} + F_{68} u_{68z} + F_{78} u_{78z} = 100

Results:

( x 5 , y 5 , z 5 ) ( 0 , 0 , 5 ) ( x 6 , y 6 , z 6 ) ( 10.95 , 0 , 6 ) ( x 7 , y 7 , z 7 ) ( 4.2 , 11.06 , 7 ) ( x 8 , y 8 , z 8 ) ( 3.14 , 0.507 , 17.6 ) F 58 72.74 F 68 30.76 F 78 5.61 (x_5, y_5, z_5) \approx (0,0,5) \\ (x_6, y_6, z_6) \approx (10.95 ,0, 6) \\ (x_7, y_7, z_7) \approx (4.2,11.06,7) \\ (x_8, y_8, z_8) \approx (3.14,0.507,17.6) \\ F_{58} \approx 72.74 \\ F_{68} \approx 30.76 \\ F_{78} \approx 5.61

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