The uniform dresser has a weight of 91 lb and rests on a tile floor for which μs = 0.25. If the man pushes on it in the horizontal direction θ = 0∘, determine the smallest magnitude of force F needed to move the dresser. Also, if the man has a weight of 151 lb , determine the smallest coefficient of static friction between his shoes and the floor so that he does not slip.

F = 22.75 lb

μ₁ = 0.15

Explanation:

The smallest force required to move the dresser must be equal to the force of friction between the man and the dresser. Therefore,

F = μR

F = μW

where,

F = Smallest force needed to move dresser = ?

μ = coefficient of static friction = 0.25

W = Weight of dresser = 91 lb

Therefore,

F = (0.25)(91 lb)

F = 22.75 lb

Now, for the coefficient of static friction between shoes and floor, we use the same formula but with the mas of the man:

F = μ₁W₁

where,

μ₁ = coefficient of static friction between shoes and floor

W₁ = Weight of man = 151 lb

Therefore,

22.75 lb = μ₁ (151 lb)

μ₁ = 22.75 lb/151 lb

μ₁ = 0.15

i'm thinking more like energy

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