Please help me solve this problem about the force swinging on a rope.?

C

CHEETAH

New member
A 45 kg child swings on a 15 m rope tied to a tree branch so that her speed at the exact bottom of the arc is 9 m/s. What is the minimum force that the child must exert on the rope without slipping?
I think the two forces holding the child are normal force and force due to gravity (so centripetal force?). Tell me if this answer is wrong:
Fn = 45 N
Fg = (45)(9.8) = 441 N
Therefore, Fn + Fg = 486.
 
The Normal force is always a RE-ACTION force and always acts PERPENDICULAR (aka Normal) to a surface. In the case of the rope and child holding it, the Normal force would act perpendicular to the rope and be a re-action to the "sqeeze force" of the hands on the rope.

The total weight of child definitely acts vertically to dislodge her from rope at the bottom of the arc.
But also, since she is in an "arc of motion" there is the centripetal force that acts (on her). At the bottom of the arc, the centripetal force acts vertically to add to her weight. So she must supply a re-action force along the rope to oppose both her weight AND the centripetal force.

Fc = centripetal force = mV²/R = (45)(9)²/15 = 243 N
Fg = girl's weight = 441 N

Minimum force (vertically) on rope = 243 + 441 = 684 N ANS
 
The Normal force is always a RE-ACTION force and always acts PERPENDICULAR (aka Normal) to a surface. In the case of the rope and child holding it, the Normal force would act perpendicular to the rope and be a re-action to the "sqeeze force" of the hands on the rope.

The total weight of child definitely acts vertically to dislodge her from rope at the bottom of the arc.
But also, since she is in an "arc of motion" there is the centripetal force that acts (on her). At the bottom of the arc, the centripetal force acts vertically to add to her weight. So she must supply a re-action force along the rope to oppose both her weight AND the centripetal force.

Fc = centripetal force = mV²/R = (45)(9)²/15 = 243 N
Fg = girl's weight = 441 N

Minimum force (vertically) on rope = 243 + 441 = 684 N ANS
 
> "Fn = 45N"

Not sure how you got that number; and I don't think it's relevant anyway. "Normal" means "perpendicular," so I don't know what you think this is perpendicular to.

Let "Fr" be the force that the child is exerting on the rope (i.e, the number we want). She's pulling down on the rope with force Fr, and (by Newton's 3rd Law) the rope is pulling UP on her with force Fr.

So the two forces acting on her (at the bottom of the arc) are:
Fg = mg (down)
Fr (up)

Therefore, the net upward force on her is:
Fr - mg

By Newton's 2nd law, the net force equals her mass times her acceleration:
Fr - mg = ma

Also, we know that whenever something is moving in a circular path, it has a centripetal acceleration of this amount:
a = v²/R
("R" is the radius of the circle; in this case, the length of the rope.)

Substitute:
Fr - mg = m(v²/R)

Solve for Fr:
Fr = m(v²/R + g)

Plug in the given values for "m", "v" and "R".
 
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