Asked by stacy
A little girl is riding a sled horizontally in the snow, the combined mass of the girl and sled is 33.4 kg. She is being pulled by her parents. Mom is pulling her with a force of 5.18 N, at an angle of 20.0 degree E of N. Dad is pulling her with a force of 11.15 N, at an angle of 33.7 degree W of N. There is a wind blowing with produces a force of 10.0 N to the east on the sled. (Use East as positive x and North as postive y)
a) What are the x and y components of the resultant force on the sled assuming the snow is frictionless?
b) What is the magnitude of the acceleration of the sled? [this one i can do if i can get a.]
c) What is the angle the acceleration of the sled makes with due North (using + for the east side and - for the west side)
a) What are the x and y components of the resultant force on the sled assuming the snow is frictionless?
b) What is the magnitude of the acceleration of the sled? [this one i can do if i can get a.]
c) What is the angle the acceleration of the sled makes with due North (using + for the east side and - for the west side)
Answers
Answered by
drwls
(a) The x (East) component of the resultant force is:
Fx = 5.18 sin 20 - 11.15 sin 33.7 + 10.0
Add them up.
The y (North) component is:
Fy = 5.18 cos 20 + 11.15 cos 33.7
(There is no wind component)
Add them up.
(b) Use the Pythagorean theorem.
F = sqrt[(Fx)^2 + (Fy)^2]
a = F/m
(c) arctan Fx/Fy
Fx = 5.18 sin 20 - 11.15 sin 33.7 + 10.0
Add them up.
The y (North) component is:
Fy = 5.18 cos 20 + 11.15 cos 33.7
(There is no wind component)
Add them up.
(b) Use the Pythagorean theorem.
F = sqrt[(Fx)^2 + (Fy)^2]
a = F/m
(c) arctan Fx/Fy
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