A belt moves over three pulleys at a constant speed v0 [ft/s]. Knowing that the normal component of the acceleration of the portion of the belt in contact with pulley B is aB,n=90 [ft/s2 ]. Determine:
(a) speed of the belt, v0 [ft/s], and
(b) normal component of the acceleration of the portion of the belt in contact with pulley A, aA,n [ft/s2 ].
Radii of pulleys are: rA =3 [in], rB =1 [in], rC=2.5 [in].
an = ||v x a|| / ||v||
The Attempt at a Solution
I believe the first step would be to find the acceleration of the belt given the normal component. Since the belt is moving at a constant speed, both the tangential component and the acceleration would be 0 ft/s2.
After that, we should put the numbers we have into the equation and get:
90 ft/s2 = ||v x 0 ft/s2|| / ||v||
..and solve for velocity. However I'm unsure how to do this since none of the numbers I'm given are in component vector form.
For part B, I would assume it is just aA,n divided by 3 since the radius of A is 3 times larger than B?