Answer:
The magnitude of the normal force is 197.7 N.
The magnitude of the kinetic frictional force is 192.5 N
Explanation:
Hi!
Please, for a better understanding of the problem, see the attached figure.
We have the following forces acting on the child (see figure):
parallel forces:
Fx = parallel component of the force exerted by the rope, F.
Fr = friction force.
wx = parallel component of the weight.
perpendicular forces:
N = normal force of the slide.
wy = perpendicular component of the weight.
Fy = perpendicular component of the force exerted by the rope, F.
The child does not have an acceleration in the perpendicular direction, then, the sum of forces acting in that direction has to be zero:
N - wy + Fy = 0 (considering upwards as positive)
Solving for N
N = wy - Fy
Using trigonometry of right triangles:
wy = W · cos 38°
wy = m · g · cos 38°
Where m = mass and g = acceleration due to gravity:
wy = 28 kg · 9.8 m/s² · cos 38° = 216.2 N
In the same way, we calculate Fy:
Fy = F · sin 38°
Fy = 30 N · sin 38°
Fy = 18.5 N
Then, the normal force will be:
N = wy - Fy
N = 216.2 N - 18.5 N = 197.7 N
The magnitude of the normal force is 197.7 N
If the child slides with constant speed, the acceleration is zero. That means that the sum of the forces acting in the direction of movement has to be zero:
Fx + wx - Fr = 0
Solving for Fr:
Fr = Fx + wx
And now we proceed in the same way as in the previous question, using trigonometry:
Fx = F · cos 38°
Fx = 30 N · cos 38° = 23.6 N
wx = W · sin 38°
wx = m · g · sin 38°
wx = 28 kg · 9.8 m/s² · sin 38°
wx = 168.9 N
Then Fr will be:
Fr = 23.6 N + 168.9 N = 192.5 N
The magnitude of the kinetic frictional force is 192.5 N
The normal force acting on the child due to child's weight is 216.2 N.
The kinetic frictional force from the child's slide is 199 N.
The given parameters;
The normal force on the child is calculated as follows;
[tex]F_n = W cos(\theta)\\\\F_n = (28\times 9.8)\times cos(38)\\\\F_n = 216.2 \ N[/tex]
The parallel force on the child is calculated as follows;
[tex]F + Wsin(\theta) - F_k = ma[/tex]
at constant speed, acceleration, a = 0
[tex]F + Wsin(\theta) -F_k = 0\\\\F_k = F + Wsin(\theta)\\\\F_k = 30 \ + \ (28\times 9.8 \times sin(38))\\\\F_k = 199 \ N[/tex]
Thus, the kinetic frictional force from the slide is 199 N.
Lear more here:https://brainly.com/question/3380489
