Asked by Hannah
1) A baby elephant is stuck in a mud hole. To help pull it out, game keepers use a rope to apply a force FA. By itself, however, force FA is insufficient. Therefore, two additional forces FB and FC are applied. Each of these additional forces has the same magnitude F. The magnitude of the resultant force acting on the elephant in part b is k times larger than that in part a. Find the ratio F/FA when k = 2.32.
The angles bewteen anad b and b and c are all 22 degrees.
I do not kno
The angles bewteen anad b and b and c are all 22 degrees.
I do not kno
Answers
Answered by
Hannah
I do not know how to solve for a ratio.
Answered by
bobpursley
Physics(thank you for the help) - bobpursley, Sunday, May 27, 2012 at 4:53pm
I am uncertain what part b is.
I am uncertain what F is in the ratio F/Fa
If F is the total force, then F/Fa=k by definition. So frankly, I have no idea what the question is asking.
Answered by
Elena
As the forces FB and FC are equal the horizontal component of force is given as FB = FC = F•cos θ,
θ =22º
The net force applied on the elephant is
Fnet = FA + 2 F•cos θ ........ (1) ( there are two forces on either side of FA)
Given that Fnet is k times FA
Fnet = k• FA where k = 2.32•FA
Fnet = 2.32• Fa .............. (2)
Plug in (2) in (1)
2.32 •FA = FA + 2• F• cos θ
2.32• FA - FA = 2• F•cos θ
(2.32 - 1) • FA = 2 F cos θ
1.32• FA= 2• F• cos θ
Ratio
F / FA = 1.32/( 2• cos θ)=
=1.32/2•cos22º = 0.71
θ =22º
The net force applied on the elephant is
Fnet = FA + 2 F•cos θ ........ (1) ( there are two forces on either side of FA)
Given that Fnet is k times FA
Fnet = k• FA where k = 2.32•FA
Fnet = 2.32• Fa .............. (2)
Plug in (2) in (1)
2.32 •FA = FA + 2• F• cos θ
2.32• FA - FA = 2• F•cos θ
(2.32 - 1) • FA = 2 F cos θ
1.32• FA= 2• F• cos θ
Ratio
F / FA = 1.32/( 2• cos θ)=
=1.32/2•cos22º = 0.71
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