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Solution 5.9

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Image:5.9.gif


As the lorry and thus all parts of the lorry have constant velocity the forces on the load must be in equilibrium.

Resolving horisontally to the right

T2cos28T1cos32=0

giving

T2=cos28cos32T1

Resolving vertically upwards.

T1cos58+T2cos62mg=0

Substituting for T2 in this equation gives an equation only containing T1.

T1cos58+cos28cos32T1cos62mg=0T1cos58+cos28cos32cos62=mgT10.53+0.45=409.8T1=3920.98=400 N

and thus using T2=cos28cos32T1

T2=0.8830.848400=384 N