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Sum of horizontal and vertical components:
The two forces shown in the figure, are to be replaced by an equivalent force R applied at P. Locate point P by finding its distance x from AB and specify magnitude of R and the angle O it makes with horizontal.
Sol.: Let us assume equivalent force R (Resultant force) is acting at an angle of θ with the horizontal. And ∑H and ∑V be the sum of horizontal and vertical components.
∑H = -1500 cos 30º = -1299N ...(i)
∑V = 1000 -1500 sin 30º = 250 N ...(ii)
R = { ∑H2 + ∑V2 }1/2
For the direction of resultant
R = 1322.87 N .......ANS
tanθ = ΣV/ ΣH
= 250/-1299
= -10.89º .......ANS
Now for finding position of resultant, we make use of Varignon's theorem, that is R × d =∑M , Take moment about point 'O'
1322.878 × x = 1500 cos 30º × 180 + 1500 sin 30º × 50 -1000 × 200
By solving x = 53.92 mm .......ANS
Thin wall round tube has diameter of 60mm. The end of it will be compressed to square for. How many percents torsion strenght will become lower at the end tube?
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