The equivalent resultant couple moment about the point O is 15.375KNm
The load carried by section (1)
F1 = 1/2 × 1.5 × (w1 - w2)
F1 = 1/2 × 1.5 × 4 = 3KN
the load carried by section (2).
F2 = w2 × (1.5 + 0.75)
F2 = 4 × 2.25 = 9KN
F3 = 1/2 × 0.75 × 4 = 1.5 KN
Determine the equivalent resultant force.
∑Fr = F1 + F2 + F3 = 13.5 KN
the moment about the point O due to distributed loading,
∑ Mr = F1 x1 + F2x2 + F3x3
x1 = 0.5m
x2 = 1.125m
x3 = 2.5m
the equivalent resultant couple moment about the point O is
Mr = ∑Mr = 15.375KNm
The equivalent resultant couple moment about any point on the given beam is equal to the sum of couple moment developed by each of the distributed loading about that point. Calculate the moment caused by each force about point O, using the distance of the centroid from end O.
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Answer:
W = 600 J
Explanation:
We have,
According to attached figure,
Height of the inclined plane is 60 m
Force acting on the block is 10 N
It is required to find the work must be done against gravity to move it to the top of the incline. The work done is given by :
W = mgh
or
So, the work done against the gravity is 600 J.
Answer:
Explanation:
Use the velocity formula to solve
In this question, you are given velocity , and you are given a distance,
. Time in this question is what you'll need to find.
Start by rearranging the velocity formula, to isolate for t.
Start by multiplying both sides by t
Then divide both sides by v.
Now that you've isolated for time, sub in your values and calculate.