Answer:
The force is [tex] F_c = 789.03 \ N [/tex]
Explanation:
From the question we are told that
The tangential resistive force is [tex]F_t = 115 \ N[/tex]
The mass of the wheel is m = 1.80 kg
The diameter of the wheel is [tex]d = 50.0 cm = 0.5 \ m[/tex]
The diameter of the sprocket is [tex]d_c = 8.50 \ cm =0.085 \ m[/tex]
The angular acceleration considered is [tex]\alpha = 4.30\ rad/s^2[/tex]
Generally the radius of the wheel is
[tex]r = \frac{d}{2}[/tex]
=> [tex]r = \frac{0.5}{2}[/tex]
=> [tex]r = 0.25 \ m [/tex]
Generally the radius of the sprocket is
[tex]r_c = \frac{d_c}{2}[/tex]
=> [tex]r_c = \frac{0.085}{2}[/tex]
=> [tex]r_c = 0.0425 \ m [/tex]
Generally the moment of inertia of the wheel is mathematically represented as
[tex]I = m * r^2[/tex]
=> [tex]I = 1.80 * 0.25^2[/tex]
=> [tex]I = 1.1125 \ kg \cdot m^2[/tex]
Generally the torque experienced by the wheel due to the forces acting on it is mathematically represented as
[tex]\tau = F_c * r_c - F_t * r [/tex]
Here [tex]F_c[/tex] is the force acting on the sprocket
So
[tex]\tau = F_c * 0.0425 - 115 * 0.25 [/tex]
[tex]\tau = 0.0425F_c - 28.75 [/tex]
Generally the torques that will cause the wheel to move with [tex]\alpha = 4.30\ rad/s^2[/tex] is mathematically represented as
[tex]\tau = I * \alpha[/tex]
So
[tex] 0.0425F_c - 28.75 = I * \alpha [/tex]
[tex] 0.0425F_c - 28.75 = 1.1125 *4.30 [/tex]
[tex] 0.0425F_c - 28.75 = 1.1125 *4.30 [/tex]
[tex] F_c = 789.03 \ N [/tex]
