Find Acceleration of Sled Pulled by Chuck with 50kg Mass

[Bdurb]

Chuck pulls his brother in a sled. Chucks brother and the sled have a combined mass of 50kg. On a flat surface, chuck pulls with 120 N 30 degrees above the horizontal. The coefficient of friction btwn the sled and the snow is .08. Find the acceleration of the sled.
i drew my FBD

but I am not sure if assumed wrong for the Fn or the normal force. Any help is greatly appreciated.

 

[thrill3rnit3]

Normal forces always acts perpendicular to the surface.

And that 500 should be a 50, based from your question.

 

[Redbelly98]

Chuck pulls his brother in a sled. Chucks brother and the sled have a combined mass of 50kg. On a flat surface, chuck pulls with 120 N 30 degrees above the horizontal. The coefficient of friction btwn the sled and the snow is .08. Find the acceleration of the sled.

i drew my FBD
http://i746.photobucket.com/albums/xx109/SubyFTW/DSC01211.jpg

but I am not sure if assumed wrong for the Fn or the normal force. Any help is greatly appreciated.

Your Fn is incorrect. Fn must act straight upward, perpendicular to the surface as thrill3rnit3 said.

Set up two equations using F_net=ma. One equation for the y-direction, and one for the x-direction.

Normal forces always acts perpendicular to the surface.

And that 500 should be a 50, based from your question.

No, the 500 is correct since he is indicating forces in Newtons.

Using g=10 m/s² gives a weight of 500 N for the 50 kg mass.

 

[thrill3rnit3]

Oops, brain fart that time :tongue:

 

[rwisz]

Thank you for providing the information about the situation. Based on the given information, we can calculate the acceleration of the sled using Newton's Second Law of Motion, which states that the net force acting on an object is equal to its mass multiplied by its acceleration (F=ma).

First, we can start by drawing a Free Body Diagram (FBD) to help us visualize the forces acting on the sled. From the given information, we can identify the following forces:

1. The force of gravity (mg) acting downwards on the sled, where m is the combined mass of the sled and Chuck's brother (50kg) and g is the acceleration due to gravity (9.8 m/s^2).

2. The normal force (Fn) acting upwards on the sled, which is equal in magnitude to the force of gravity. This is because the sled is on a flat surface and there is no vertical acceleration.

3. The force of friction (Ff) acting in the opposite direction of motion, which is equal to the coefficient of friction (μ) multiplied by the normal force (Ff=μFn). In this case, μ=0.08 and Fn=mg.

4. The force of Chuck pulling the sled (Fp), which has a magnitude of 120 N and is at an angle of 30 degrees above the horizontal.

Now, we can use the FBD to set up our equation:

ΣF=ma

Where ΣF is the net force acting on the sled and a is the acceleration we are trying to find.

ΣF= Fp - Ff

We can break down the components of Fp into its x and y components:

Fx=120cos30=103.9 N
Fy=120sin30=60 N

Substituting these values into our equation, we get:

ΣF=103.9 N - (0.08)(50 kg)(9.8 m/s^2)= 59.2 N

Now, we can solve for the acceleration:

ΣF=ma
59.2 N= (50 kg)a
a= 1.18 m/s^2

Therefore, the acceleration of the sled is 1.18 m/s^2 in the direction of Chuck's pull. I hope this helps! Please let me know if you have any further questions.