Given :
Force, .
Force is acting at point A( 2 m, 3 m ) and B( 3 m, 5 m )
To Find :
The work done by force F .
Solution :
Displacement vector between point A and B is :
Now, we know work done is given by :
W = 12000 J
Therefore, work done by force is 12000 J .
Answer:
work done by electric field is 0.06 J
Explanation:
Given data:
Two point charge is
0+1 charge positioned is (0 cm , 1 cm, 0.00 cm)
-1 charge positioned is (0 cm , -1 cm, 0.00 cm)
E = 3.0\times 10^6 N/C
From above information, the distance between given two charges d = 2 cm
then d = 0.02m
work needed is W = q E d
W = 0.06 J
Therefore work done by electric field is 0.06 J
Answer:
a) Fb= 275.77 lb Fc= 142.75 lb
b) M = -779.97 lb.ft (i.e. 779.97 lb.ft in clockwise direction)
c) Fax = 195 lb
Fay = 337.75 lb
Fbx = 195 lb
Fby = 195 lb
Explanation:
Question: Three tugboats are used to turn a barge in a narrow channel. To avoid producing any net translation of the barge, the forces applied should be couples. The tugboat at point A applies a 390 lb force.
(a) Determine FB and FC so that only couples are applied.
(b) Using your answers to Part (a), determine the resultant couple moment that is produced.
(c) Resolve the forces at A and B into x and y components, and identify the pairs of forces that constitute couples.
Solution:
<u>For this problem Right hand side is positive X direction and Upwards is positive Y direction. Couples and moments will be considered positive in counterclockwise direction.</u>
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a) For no translation condition
∑ & ∑
Hence,
and
Inserting the value of and solving the remaining equations simultaneously yields (magnitudes),
b) Summing up moments
(i.e. 779.97 lb.ft clockwise)
c)
