AP PROBLEM CHILD ON A SWING

AN ADULT EXERTS A HORIZONTAL FORCE ON A SWING THAT IS SUSPENDED BY A ROPE OF LENGTH L, HOLDING IT AN AN ANGLE q WITH THE VERTICAL. THE CHILD IN THE WING HAS WEIGHT W AND DIMENSIONS THAT ARE NEGLIGIBLE COMPARED TO L. THE WEIGHTS OF THE ROPE AND OF THE SEAT ARE NEGLIGIBLE. IN TERMS OF W AND q DETERMINE



A) THE TENSION IN THE ROPE

B) THE HORIZONTAL FORCE EXERTED BY THE ADULT

THE ADULT RELEASE THE SWING FROM REST. IN TERMS OF W AND q, DETERMINE

C) THE TENSION IN THE ROPE JUST AFTER THE RELEASE (THE SWING IS INSTANTANEOUSLY AT REST)

D) THE TENSION IN THE ROPE AS THE SWING PASSES THROUGH IS LOWEST POINT

A) T = W/cos(q) (b/c Tcos(q) = W)

B) F = W*tan(q) (b/c F = Tsin(q)

C) T = W*cos(q) (Find T so that acceleration is tangential to the circle)

D) Not sure but if you want it in terms of W and q I used potential and kinetic energy in terms of L and then plugged it into my equation T = W + (W/10)(v^2/L). The L's cancel and I came up with T = W + 2W*cos(q). (note: i used g = 10 m/s^2)

To solve this problem, we can analyze the forces acting on the swing at different positions.

A) The tension in the rope can be found by considering the vertical forces. At rest, the tension in the rope equals the weight of the child, W. However, when the swing is at an angle q with the vertical, the tension in the rope can be resolved into two components: Tcos(q) acting horizontally (to balance the adult's force) and Tsin(q) acting vertically (to balance the child's weight).

Tension in the rope, T = W / cos(q)

B) The horizontal force exerted by the adult equals the tension in the rope multiplied by sin(q), because sin(q) represents the vertical component of the tension.

Horizontal force exerted by the adult, F = T * sin(q) = (W / cos(q)) * sin(q)

C) When the adult releases the swing and it reaches rest again, the tension in the rope just after the release is equal to the weight of the child, W.

Tension in the rope just after the release = W

D) The tension in the rope at the lowest point can be determined by considering the forces acting at that position. At the lowest point, the swing's weight and the tension in the rope act in the same direction. Therefore, the tension in the rope at the lowest point is the sum of the weight of the child, W, and the weight of the seat, which is negligible.

Tension in the rope at the lowest point = W + 0 (negligible seat weight) = W

So, in summary:

A) Tension in the rope, T = W / cos(q)
B) Horizontal force exerted by the adult, F = (W / cos(q)) * sin(q)
C) Tension in the rope just after the release = W
D) Tension in the rope at the lowest point = W

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