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Evgesh-ka [11]
2 years ago
9

A car has a steering wheel with a 15 inch diameter that takes 18 lbs of Effort force to move is

Engineering
1 answer:
elena-14-01-66 [18.8K]2 years ago
4 0

Answer: A first class lever in static equilibrium has a 50lb resistance force and 15lb effort force. The lever's effort force is located 4 ft from the fulcrum.

Explanation:

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Answer:

  use the percentage error relation

Explanation:

The percentage error in anything is computed from ...

  %error = ((measured value)/(accurate value) -1) × 100%

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The difficulty with voltage measurements is that the "accurate value" may be hard to determine. It can be computed from the nominal values of circuit components, but there is no guarantee that the components actually have those values.

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A cantilever timber beam with a span of L = 4.25 m supports a linearly distributed load with maximum intensity of w0 = 5.5 kN/m.
9966 [12]

Answer:

the minimum width is b= 0.1414m = 141mm

Explanation:

]given,

L= 4.25

w₀ = 5.5kN/m,

allowable bending stress = 7MPa

allowable shear stress = 875kPa

h/b = 0.67

b = ?

for a linearly distributed load, with maximum load intensity, w₀ of 5.5kN/m,

the maximum moment, M exerted by the timber is =  \frac{w₀ L²}{9√3}[/texM = w₀ L²}/{9√3 = 99.34/15.6 =6.367kNmfor a linearly distributed load, with maximum load intensity, w₀ of 5.5kN/m, the shear force, V =  [tex]\frac{w₀ L}{2}[/texV = {w₀ L}/{3} = 7.79kNfor maximum bending stress of a rectangular timber, B, = [tex]\frac{6M}{bh²}

given h/b = 0.67, i.e h=0.67b

allowable bending stress = \frac{6M}{bh²} = 7000kPa

7000  = (6*6.37)/ (b *(0.67b)² ) = 38.21/0.449b³

3080b³=38.21

b³ = 38.21/3080 = 0.0124

b = 0.232m

h=0.67b = 0.67* 0.232 = 0.155m

for allowable  shear stress = (3V)/(2bh)

875 = (3V)/(2bh) = (3x 7.79)/(2xbx0.67b)

875 = 23.375/1.34b²

1172.5 b²= 23.375

b² =0.0199

b= 0.1414m

h=0.67b = 0.67* 0.1414 = 0.095m

the minimum width is b= 0.1414m = 141mm

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