Asmall box is held in place against a rough vertical wall bysomeone pushing on it with a force directed upward at 28 degreesabove horizontal. the coeffients of static and kineticfriction between the box and the wall are 0.40 and 0.30,respectively. the box slides down unless the applied forcehas a magnitude 23n. what is the mass of the box?

m = 1.9287 Kg

Explanation:

Given

∅ = 28º

μs = 0.40

μk = 0.30

F = 23 N

m = ?

We can apply  

∑Fh = 0  (+→)

N - Fh = 0    ⇒  N = Fh = F*Cos ∅

then    N = F*Cos ∅

∑Fv = 0  (+↑)

⇒    Fy - W + Ff = 0

where

Fy = F*Sin ∅

W = m*g

Ff = μs*N = μs*F*Cos ∅

then

⇒    Fy - W + Ff = F*Sin ∅ - m*g + μs*F*Cos ∅ = 0

⇒    m = (F / g)*(Sin ∅ + μs*Cos ∅)

⇒    m = (23 N / 9.81 m/s²)*(Sin 28º + 0.40*Cos 28º)

⇒    m = 1.9287 Kg

  the unit of the measure =

explanation:

  the property that is quantified by the energy needed per of material to raise the temperature by one.

  si unit of energy = joule (j)

  si unit of temperature = kelvin (k)

  si unit of mass of material = kilogram (kg)

the property that is quantified by the energy needed per of material to raise the temperature by one is called specific heat.

the unit of specific heat is

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west