Find the magnitudes of all of the forces acting on the sled

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A person is analyzing the forces acting on a 5 kg sled being pulled at an angle of 30 degrees, resulting in horizontal acceleration of 1.7 m/s² without friction. The equations of motion are applied, leading to the calculation of the normal force and the tension in the rope. The user initially calculates the magnitudes of the forces but finds that their sum does not equal the expected net force of 8.5 N. A key point of clarification is that forces must be added as vectors, considering their components rather than as scalars. The discussion emphasizes the importance of correctly applying vector addition to resolve the forces acting on the sled.
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Homework Statement


A person pulls on a rope attached to a 5 kg sled. The rope makes an angle of 30 degrees with the horizontal. Sled accelerates horizontally at 1.7 m/s^2. There is no friction.

Homework Equations


ma = Fnet
mg = Fgravity

The Attempt at a Solution


Fnorm = Earth force up on sled
Fr = the force of the rope of the sled being pulled

The x and y components for Fr would be y = Frsinθ and x=Frcosθ.
Vector of Fnorm = <0, Fnorm>
Vector of Fg = <0, -50>
Vector of Fr = <Frcos30, Frsin30>

Fnetx => 8.5 = Frcos30
Fnety => 0 = Fc - 50 + Frsin30

So I got this far and starting doing algebraic stuff to solve for Fr and then plugging that into the second equation. From there, I solved for Fc, and the value was 45.095 N. And from the Fnetx equation, Fr = 9.8 N.
However, when I added all the magnitudes of forces, my answer does not equal to the Fnet, which is ma = 8.5 N.
Can someone help me with this? Have I forgotten something? Am I even doing this right?
 
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Show us how you added up the forces to get the net force. Your answers so far look OK.

Chet
 
For the net force, I did the force magnitude of Fnorm plus force magnitude of Fr minus the force magnitude of Fg, which is 8.5 N + 45.095 N - 50 N. That answer got me 3.595 N.
 
NkaujHelp said:
For the net force, I did the force magnitude of Fnorm plus force magnitude of Fr minus the force magnitude of Fg, which is 8.5 N + 45.095 N - 50 N. That answer got me 3.595 N.
The forces have to be added as vectors, not as scalars.

Chet
 
Chestermiller said:
The forces have to be added as vectors, not as scalars.

Chet

How do I do that?
 
NkaujHelp said:
How do I do that?
You do it by adding their components. But, you've already done that.

Chet
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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