2nd Condition of Equilibrium... *Given, Required, Solution
1.) A uniform bar 4 meters long weighs 90 N. A 120 N weight is hung at one end of the bar. How far from the free end must the bar be supported for it to remain in horizontal?
2.) The center of gravity of a 100 grams meter stick is located at 50.5 and the stick is supported at 60 cm. Where must a 190 grams load be hung in order to have equilibrium?
1.) A uniform bar 4 meters long weighs 90 N. A 120 N weight is hung at one end of the bar. How far from the free end must the bar be supported for it to remain in horizontal?
2.) The center of gravity of a 100 grams meter stick is located at 50.5 and the stick is supported at 60 cm. Where must a 190 grams load be hung in order to have equilibrium?
Answers & Comments
Answer:
1.. The minimum length of the wrench will assume that the maximum force is applied at an angle of 90∘. Therefore, we can use the simplified expression for torque:
T=F⋅r
Here, r is the length of the wrench.
Rearranging for length and plugging in our values, we get:
r=TF=15N⋅m40N=0.375m. 2.meter stick is located at 50.5 cm and the stick is supported at 60 cm. where must a 190 gram load be hung in order to have equilibrium?
Explanation:
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