Problem 3- A 40 kg child swings in a swing supported by two ropes, each 3 m long. (a) If the tension in each rope at the lowest point is 350 N,
a. find the child's speed at the lowest point.
b. the force exerted by the seat on the child at the lowest point. (Neglect the mass of
the seat.)
(b) If the tension in each rope is 212 N when the ropes make an angle 30 degree with the
vertical, find the total acceleration of the child.
tensionbottom=mg+mv^2/r
solve for v.
force on set must equal tension...
b.total acceleration is a vector of g downward, and mv^2/r along the radial.
but tension alone, along the radial,
tension=mgcosTheta+mv^2/r
but tension and gravity force=net force
net force=mass*accelerationnet
now you have a vector equation, net force is in the direction given by adding tension along that direction, and mgcosThea along the radial, and mgsinTheta downward. So solve for that net force, then net force=mass*acceleration and solve for acceleration. Not a first grade problem.
بسج غش المره اليايه حلي بروحج فهمتي
ما عليج منها هذي تغار منج
To solve these problems, we can use principles of circular motion and Newton's laws of motion.
(a) To find the child's speed at the lowest point, we can use the equation for centripetal force:
Fc = mv^2/r
Where Fc is the centripetal force, m is the mass of the child, v is the velocity, and r is the radius (length of the rope).
Given that the tension in each rope at the lowest point is 350 N, we can assume this is the centripetal force acting on the child:
Fc = 350 N
The tension in each rope provides the centripetal force, so we have:
2T = Fc
Where T is the tension in each rope.
We can substitute Fc and rearrange the equation to solve for v:
mv^2/r = 2T
v^2 = 2Tr/m
v = sqrt(2Tr/m)
Substituting the given values, we have:
v = sqrt(2 * 350 N * 3 m / 40 kg)
v ≈ 4.08 m/s
Therefore, the child's speed at the lowest point is approximately 4.08 m/s.
(b) To find the force exerted by the seat on the child at the lowest point, we can use Newton's second law of motion:
F = ma
Since the child is at the lowest point, the total force acting on the child is the sum of the tension in each rope and the gravitational force:
F = T + mg
Where m is the mass of the child, g is the acceleration due to gravity, and T is the tension in each rope.
Substituting the given values, we have:
F = 2T + (40 kg)(9.8 m/s^2)
F = 2(350 N) + (40 kg)(9.8 m/s^2)
F ≈ 1372 N
Therefore, the force exerted by the seat on the child at the lowest point is approximately 1372 N.
(b) To find the total acceleration of the child when the tension in each rope is 212 N and the ropes make a 30-degree angle with the vertical, we can resolve the forces acting on the child.
At this angle, the vertical component of the tension in each rope counteracts the gravitational force, while the horizontal component provides the centripetal force.
The vertical component of tension is given by:
T * cos(30°)
The horizontal component of tension is given by:
T * sin(30°)
The net vertical force acting on the child is:
F_vert = T * cos(30°) - m * g
The net horizontal force acting on the child is given by the centripetal force:
F_hori = m * a_c
Where m is the mass of the child, g is the acceleration due to gravity, and a_c is the centripetal acceleration.
Since the circular motion is due to the horizontal component of tension, we can equate the net horizontal force to the centripetal force:
F_hori = Fc
m * a_c = T * sin(30°)
Rearranging the equation, we have:
a_c = T * sin(30°) / m
Substituting the given values, we have:
a_c = (212 N) * sin(30°) / (40 kg)
a_c ≈ 2.9 m/s^2
Therefore, the total acceleration of the child is approximately 2.9 m/s^2.