Why Do My Calculations on Frictionless Hill Physics Seem Incorrect?

[kimikims]

I tried working this problem- but it doesn't come out right.. any help? My answers are in blue.

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A child holds a sled on a frictionless, snowcovered hill, inclinded at an angle of 27 degrees.

1) If the sled weights 60 N, find the force exerted on the rope by the child. Answer in units of N.

90 degrees - 27 degrees = 63 degrees
cos 63 = 60/x
x cos 63 = 60
x = 60/cos 63
x = 132.2 N

2) What force is exerted on the sled by the hill? Answer in units of N.

132.2 N tan 27 = 67.35

 

[Sirus]

Consider all the forces acting on the sled (draw a free-body diagram). The vertical gravitational force is not the only one...

 

[wais]

N

Hi there! It looks like you may have made a couple of errors in your calculations. Here's how I would approach this problem:

1) To find the force exerted on the rope by the child, we can use the formula F = mg sinθ, where F is the force, m is the mass, g is the acceleration due to gravity, and θ is the angle of inclination. In this case, we know that the mass of the sled is 60 N and the angle of inclination is 27 degrees. So we can plug those values in and get:

F = (60 N)(9.8 m/s^2)sin(27 degrees) = 153.4 N

So the force exerted on the rope by the child is 153.4 N.

2) To find the force exerted on the sled by the hill, we can use the formula F = mg cosθ. Again, we know that the mass of the sled is 60 N and the angle of inclination is 27 degrees. So we can plug those values in and get:

F = (60 N)(9.8 m/s^2)cos(27 degrees) = 132.2 N

So the force exerted on the sled by the hill is 132.2 N.

I hope this helps! It's always a good idea to double check your calculations and make sure you're using the correct formulas. Good luck with your problem!