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The bearing pin shown in (Figure 1) supports the load of P = 3.5 kN. The support is 12 mm thick. Part A Determine the normal

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N im.Z 3. Efx=0 N- Pcos 30 = 0 > N= 3.s cosso N= 3.031 kN = 30310 Efyao v- P sin 30 =0 . » V- 3.5 sin 30 v=1.75 kN = 1750N (

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