A small sphere with mass 5.00.×10−7kg and charge +3.00μC is released from rest a distance of 0.600 m above a large horizontal in
2 answers:
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Answer:
- the forces on the left hand side is 1.038 kN
- the forces on the right hand side is 1.483 kN
Explanation:
Given the data in the question, as illustrated in the image below;
Length of the scaffold = 3 m
weight of the scaffold = 395 N
Weight of Bob = 805 N and stands 1 m from the left end
weight of washing equipment = 500N and on sits 2 m from the left end
Weight of Joe = 820 N and stand 0.500 m from the right end
so the force on the left cable will be;
=
[ (805 N)( (3-1) m) + ( 395 N )(
) + ( 500 N )(1m ) + ( 820 N)( 0.500m ) ]
=
[ 1610 + 592.5 + 500 + 410 ]
=
[ 3112.5 ]
= 1037.5 N
= 1.038 kN
Therefore, the forces on the left hand side is 1.038 kN
On the right hand side;
= ( 805 N + 395 N + 500 N + 820 N ) - 1037.5 N
= 2520 N - 1037.5 N
= 1482.5 N
= 1.483 kN
Therefore, the forces on the right hand side is 1.483 kN
