Answer:
33.4
Explanation:
Step 1:
\sumMo=0 (moment about the origin)
Fb(15)-Fc(15)=0
Fb=Fc
Step 2:
\sumFx=0
-Fb-Fccos\theta+Ncsin\theta=0
Fc=0.3Nc=Fb
-0.3Nc-0.3Nccos\theta+Ncsin\theta=0
(-0.3-cos\theta+sin\theta)Nc=0----(1)
\sumFy=0
Nccos\theta+Fcsin\theta-Nb=0
Nccos\theta+0.3Ncsin\theta-Nc=0
Nc[cos\theta+0.3sin\theta-1]=0--------(2)
Solving eq (1) and eq (2)
\theta=33.4
Step 3:
As the roller is a two force member
2(90-\phi)+\theta=180
\phi=\theta/2
\phi=Tan(\muN/N)-1
\phi=16.7
\theta=2x16.7=33.4
Answer:
this might help
Explanation:
https://science.ksc.nasa.gov/mars/msp98/misc/MCO_MIB_Report.pdf
Answer:
They help an organization realize the maximum benefit from its investment in equipment, personnel, and business processes. They use and analyze systems, interpret data, and customize systems to better meet the organization's needs.
Explanation:
brainliest plz
Answer:
Explanation:
For pressure gage we can determine this by saying:
The closed tank with oil and air has a pressure of P₁ and the pressure of oil at a certain height in the U-tube on mercury is p₁gh₁. The pressure of mercury on the air in pressure gauge is p₂gh₂. The pressure of the gage is P₂.
We want to work out P₁-P₂: Heights aren't given so we can solve it in terms of height: assuming h₁=h₂=h
