The slotted arm OA rotates about a fixed axis through O. At the instant under consideration, θ = 36°, θ˙ = 36 deg/s, and θ¨ = 26 deg/s2. Determine the magnitude of the force F applied by arm OA and the magnitude of the force N applied by the sides of the slot to the 0.5-kg slider B. Neglect all friction, and let L = 0.72 m. The motion occurs in a vertical plane.

   F = 0.0545 N

Explanation:

Let's use Newton's second law for rotational movement

        τ = I α

The moment of inertia for a bar supported by some ends is

       I = 1/3 m L²

Torque is

            τ = F L

Let's replace

         F L = 1/3 m L² α

         F = 1/3 m L α

Let's reduce angular acceleration to SI units

       Alf = 26º / s² (π rad / 180º) = 0.454 rad / s²

Let's calculate

          F = 1/3  0.5  0.72  0.454

          F = 0.0545 N

the answer is b ) square root of 5

force is mass*acceleration.

we have a very small mass: 11.4*0.001 kg

if we want 0.035n of force, we need an acceleration:

0.035 = 11.4*0.001*a

a = 0.035/(11.4*0.001) m/s^2