Answer:
Normal force = 0.326N
Explanation:
Given that:
mass released from rest at C = 3.7 g = 3.7 × 10⁻³ kg
height of the mass = 1.1 m
radius = 0.2 m
acceleration due to gravity = 9.8 m/s²
We are to determine the normal force pressing on the track at A.
To to that;
Let consider the conservation of energy relation; which says:
mgh = mgr + 1/2 mv²
gh = gr + 1/2 v²
gh - gr = 1/2v²
g(h-r) = 1/2v²
v² = 2g(h-r)
However; the normal force will result to a centripetal force; as such, using the relation
N =mv²/r
replacing the value for v² = 2g(h-r) in the above relation; we have:
Normal force = 2mg(h-r)/r
Normal force = 2 × 3.7 × 10⁻³ × 9.8 ( 1.1 - 0.2 )/ 0.2
Normal force = 0.065268/0.2
Normal force = 0.32634 N
Normal force = 0.326N
Boussinesq's influence factor is high Westergaard's theory:- 1. ) Assumes that the soil medium is anisotropic 2) Deals with thin sheets of rigid material sandwiched in a homogeneous medium. ... 5) Westergaard's influence factor is low compare to Boussinesq's influence factor.
