The spring-held follower AB has weight W and moves back and forth as its end rolls on the contoured surface of the cam, where the radius is r and z = asin(2θ). If the cam is rotating at a constant rate θ', determine the force at the end A of the follower when θ = θ 1. In this position the spring is compressed δ1. Neglect friction at the bearing C.

some parts of your question is missing attached below is the missing part

Answer :  Fa = 4.46 Ib

Explanation:

use the equation

Z = 0.1 sin2∅

next we will differentiate the equation to get the locus of the velocity

z = 0.2 cos2∅∅

differentiate the equation furthermore to get the locus of acceleration in the horizontal axis

z = -0.4sin2∅(∅)^2 + 0.2cos2∅∅

note : express ∅ as 6 rad. for angular velocity and ∅ = 0 for angular acceleration

equation above becomes :

Z = - 0.4 sin 2∅ ( 6)^2 + 0.2 cos 2∅(0)

  = - 14.4 sin 2∅ ( acceleration of the follower in horizontal direction )

next calculate The force at the end of A of the follower

Fa - Kx = mz  

note: m = w / g hence : Fa - Kx = w/g z  ------- (2)

w = weight of the spring-held follower = 0.75 Ib

x = compression of the spring = 0.4

k = spring stiffness = 12 Ib/ft

∅ = 45⁰

g = 32.2 ft/s^2

input these values into equation 2

hence : Fa = 4.46 Ib ( force at the end A of the follower )

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explanation: