Illinois furniture , Inc produces all types of coffee furniture the executive secretary is a chair that has been designed using
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Answer:
A single force, which is acting at angle θ from a horizontal axis, can be resolved into components which act along the perpendicular axis.
Consider the perpendicular axis x and y, where x represents the horizontal axis and y represents vertical axis.
The Force is resolved into 2 parts, one acts along x-axis and is represent by X. The other acts along y-axis and is represented by Y.
From the diagram we can see that the Force and its components X and Y makes up a right angles triangle, where θ is the angle from the x-axis
<h3 /><h3>Find X:</h3>We know that:
cosθ = Base/Hypotenuse
cosθ = X/F
X = Fcosθ
<h3>Find Y:</h3>
We know that:
sinθ = Perpendicular/Hypotenuse
sinθ = Y/F
Y = Fsinθ
<h3>Relation of Force and its Components:</h3>
Force F can be represent by:
F = Fcosθ (along x-axis) + Fsinθ (along y-axis)
As they form a right angled triangle, we can use Pythagoras Theorem to show the relation between Force and its components.
Hypotenuse² = Base² + Perpendicular²
F² = X² + Y²
F² = (Fcosθ)² + (Fsinθ)²
Where θ can be found by using any of the trignometric functions.
Answer:
total resistance = 0.18414 K/W
Explanation:
given data
length L = 8 mm
siding = 40 mm
siding = 100 mm
studs = 0.65-m
paper faced, 28 kg/m³
gypsum layer = 12-mm
to find out
thermal resistance associated with a wall that is 2.5 m high by 6.5 m wide (having 10 studs, each 2.5 m high)
solution
we will apply here resistance formula that is
resistance = ...................1
here L is length and Ka is thermal conductivity and A id area
thermal conductivity of hard wood siding = 0.94 W/m-K
and thermal conductivity of hard wood stud = 0.16 W/m-K
and thermal conductivity of glass fiber insulation = 0.038 W/m-K
and thermal conductivity of gypsum wall board = 0.17 W/m-K
so resistance for wood is
resistance Rw = = 0.0549 K/W ................2
and resistance for stud is
resistance Rs = = 6.25 K/W ....................3
and resistance for insulation
resistance Ri = = 1.7256 K/W .................4
and resistance for wall board
resistance Rg = = 0.4343 K/W .................5
so here stud and insulated are parallel
so resistance =
we get resistance = = 1.3522 K/W ..........................6
so total resistance is
total resistance add equation 2 and equation 5 and 6
total resistance = 0.0549 + 0.4343 + 1.3522
total resistance = 1.8414 K/W
and
studs are 10
so total resistance will be
total resistance =
total resistance = 0.18414 K/W