Answer:
- F1.x ≈ -28.93
- F1.y ≈ 34.47
- F2.x = 70
- F2.y = 0
- (F1+F2).x ≈ 41.07
- (F1+F2).y ≈ 34.47
- |F1+F2| ≈ 53.62
- ∠(F1+F2) ≈ 40.0°
Explanation:
A suitable calculator can show you the vector components and their resultant in polar or rectangular format. (See attached.) 2D vectors are conveniently treated as complex numbers, which is why the y-component values are shown as imaginary.
(The 50° angle measured from the -x axis is equivalent to 130° measured from the +x axis, which is the reference we're using here.)
If you'd like to compute the vector components by hand, they are ...
(x, y) = magnitude×(cos(angle), sin(angle))
This notation is sometimes abbreviated <em>magnitude cis angle</em>, a reference to the complex number form x+yi.
Moisture content is measured in terms of pounds of water per pound of air (lb water/lb air) or grains of water per pound of air (gr. of water/lb air).
Hope this helps❤
Answer:
The Estimated uncertainty in a nominal displacement of 2 cm at the design stage is plus or minus 0.0124cm
Explanation:
uncertainty in a nominal displacement
= (u^2 + v^2)^(1/2)
assume from specifications that k = 5v/5cm
= 1v/cm
u^2 = (0.0025*2)^(2) + (0.005*10*2)^2 + (0.0025*2)^2
= 0.01225v
v = 2v * 0.001
= 0.002v
uncertainty in a nominal displacement
= (u^2 + v^2)^(1/2)
= ((0.01225)^2 + (0.002)^2)^(1/2)
= 0.0124 cm
Therefore, The Estimated uncertainty in a nominal displacement of 2 cm at the design stage is plus or minus 0.0124cm
Complete Question:
A metal plate of 400 mm in length, 200mm in width and 30 mm in depth is to be machined by orthogonal cutting using a tool of width 5mm and a depth of cut of 0.5 mm. Estimate the minimum time required to reduce the depth of the plate by 20 mm if the tool moves at 400 mm per second.
Answer:
26 mins 40 secs
Explanation:
Reduction in depth, Δd = 20 mm
Depth of cut, 
Number of passes necessary for this reduction, 
n = 20/0.5
n = 40 passes
Tool width, w = 5 mm
Width of metal plate, W = 200 mm
For a reduction in the depth per pass, tool will travel W/w = 200/5 = 40 times
Speed of tool, v = 100 mm/s
minimum time required to reduce the depth of the plate by 20 mm:
number of passes * Time/pass
n * Time/pass
40 * 40
1600 = 26 mins 40 secs
Answer:
r=0.31
Ф=18.03°
Explanation:
Given that
Diameter of bar before cutting = 75 mm
Diameter of bar after cutting = 73 mm
Mean diameter of bar d= (75+73)/2=74 mm
Mean length of uncut chip = πd
Mean length of uncut chip = π x 74 =232.45 mm
So cutting ratio r
r=0.31
So the cutting ratio is 0.31.
As we know that shear angle given as
Now by putting the values
\
Ф=18.03°
So the shear angle is 18.03°.
