Answer:
Given:
mass of robot, m = 6 kg
radius of the wheel, r = 20 mm = 0.02 m
acceleration, a = 10 [tex]mm/s^{2} = 10\times 10^{-3} m/s^{2}[/tex]
acceleration due to gravity, g = 9.8[tex]m/s^{2}[/tex]
inclination, [tex]\theta = 45^{\circ}[/tex]
Solution:
Refer to fig 1
Force is given by:
F = mgsin[tex]\theta[/tex]
Torque is given by:
Torque, [tex]\tau = F\times r = mgrsin\theta [/tex]
[tex]\tau = 6\times 9.8\times 0.02\times sin45 [/tex]
[tex]\tau = 6\times 9.8\times 0.02\times \frac{1}{\sqrt {2}}[/tex]
[tex]\tau = 0.83 N-m[/tex]
Refer to fig 2.
Provided the acceleration of the wheel, the Force and torque is now:
Force, F = mgsin[tex]\theta + ma[/tex]
Torque, [tex]\tau = F\times r = Mgrsin\theta + Mar[/tex]
[tex]\tau = 6\times 9.8\times 0.02\times \frac{1}{\sqrt {2}} + 6\times 10\times 10^{-3}\times 0.02[/tex]
[tex]\tau = 0.8312 N-m[/tex]
