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A fisherman has caught a fish on his pole, and to keep the pole steady he has to apply an upward force of F2 = 240 N at an angle of 83.5° with respect to the pole (see figure). The length of his pole is 4.1 m, and he is holding it a distance 0.35 m from the end, where he is applying a downward force F1.
With how much force, F1, in newtons, does he have to push straight downward on the end of his pole to keep the pole from moving? You may assume the pole is massless. F1 = What is the mass of the fish on the end of the pole, in kilograms? m =

Answer :

Given that,

F2 = 240N

Angle = 83.5 degree

Length = 4.1m

Distance = 0.35m

How much force, F1 in newtons ?

What is the mass of the fish on the end of the pole ?

Thus,

Calculating Torque about ‘A’

For equillibrium ∑τa= 0

F1 * L + F2 * (L – d) = 0

F1 * L= - F2* (L – d)

F1L SinΘ= F2(L – d) SinΘ

F1 = F2 (L – d)/L

F1 = 240 (4.1 – 0.35)/4.1

F1 = 219.51N

For equillibrium ∑ Fnet = 0

F2 + F1 + mg = 0

mg = F2- F1

m = F2 - F1/ g

m = 2.090Kg

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