Refer to the diagram shown below.
The basket is represented by a weightless rigid beam of length 0.78 m.
The x-coordinate is measured from the left end of the basket.
The mass at x=0 is 2*0.55 = 1.1 kg.
The weight acting at x = 0 is W₁ = 1.1*9.8 = 10.78 N
The mass near the right end is 1.8 kg.
Its weight is W₂ = 1.8*9.8 = 17.64 N
The fulcrum is in the middle of the basket, therefore its location is
x = 0.78/2 = 0.39 m.
For equilibrium, the sum of moments about the fulcrum is zero.
Therefore
(10.78 N)*(0.39 m) - (17.64 N)*(x-0.39 m) = 0
4.2042 - 17.64x + 6.8796 = 0
-17.64x = -11.0838
x = 0.6283 m
Answer: 0.63 m from the left end.
Answer:
a) 12.8212 N
b) 12.642 N
Explanation:
Mass of bucket = m = 0.54 kg
Rate of filling with sand = 56.0 g/ sec = 0.056 kg/s
Speed of sand = 3.2 m/s
g= 9.8 m/sec2
<u>Condition (a);</u>
Mass of sand = Ms = 0.75 kg
So total mass becomes = bucket mass + sand mass = 0.54 +0.75=1.29 kg
== > total weight = 1.29 × 9.8 = 12.642 N
Now impact of sand = rate of filling × velocity = 0.056 × 3.2 = 0.1792 kg. m /sec2=0.1792 N
Scale reading is sum of impact of sand and weight force ;
i-e
scale reading = 12.642 N+0.1792 N = 12.8212 N
<u>Codition (b);</u>
bucket mass + sand mass = 0.54 +0.75=1.29 kg
==>weight = mg = 1.29 × 9.8 = 12.642 N (readily calculated above as well)
Answer:
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Explanation:
Coefficients is your answer. I hope I helped:)