A sphere has a spherically symmetric mass distribution. Its average density is 3850 kg/m3. The sphere rolls and/or slides down the incline as shown below. The force which the incline applies on the sphere can be divided into two components, one parallel to the incline (called friction) and the other perpendicular to the incline (called the normal force). Calculate the size of the Normal force on the sphere. Note that the incline passes through at least 2 grid intersections and the sphere is tangent to vertical grid lines.
http://loncapa.hep.uprm.edu/cgi-bin/plot.png?file=802084083_uprm_1347075781_29709268_plot.data
http://loncapa.hep.uprm.edu/cgi-bin/plot.png?file=802084083_uprm_1347075781_29709268_plot.data
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By M = V x d
=>M = 4/3πr^3 x d
=>M = 4/3 x 3.14 x (7.62 x 10^-2)^3 x 3850
=>M = 7.13 kg
Thus Normal force (N) = Mgcosθ
=>N = 7.13 x 9.8 x [11/√{(3.25)^2 + (11)^2}]
=>N = 67.01 Newton
=>M = 4/3πr^3 x d
=>M = 4/3 x 3.14 x (7.62 x 10^-2)^3 x 3850
=>M = 7.13 kg
Thus Normal force (N) = Mgcosθ
=>N = 7.13 x 9.8 x [11/√{(3.25)^2 + (11)^2}]
=>N = 67.01 Newton