Calculating Forces for a Sled on an Inclined Hill

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Homework Help Overview

The discussion revolves around calculating forces acting on a sled on a frictionless inclined hill, specifically focusing on the force exerted on the rope by a child and the force exerted on the sled by the hill. The sled's weight is given as 73N, and the incline is at a 34-degree angle.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the relationship between the forces acting on the sled and the rope, questioning the effects of the incline angle and the components of gravitational force. There are discussions about drawing free body diagrams and calculating forces using trigonometric functions.

Discussion Status

Participants are actively engaging with the problem, sharing their calculations and questioning their understanding of the forces involved. Some have suggested drawing diagrams and breaking down the forces into components, while others express confusion about the angles and the role of gravity.

Contextual Notes

There is an emphasis on understanding the components of gravitational force acting on the sled, with some participants noting the need to clarify which angles correspond to the components of gravity. The problem's constraints include the frictionless nature of the hill and the specific weights and angles provided.

anglum
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[SOLVED] Force Exerted?

A child holds a sled on a frictionless snow covered hill, inclined at a 34 degree angle

if the sled weighs 73N find the force exerted on the rope by the child. answer in units of N
and find the force exerted on the sled by the hill, answer in units of N
 
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the force on the rope would be 73N to equal out the 73N of the sled

and the force the hill exerts on the sled would be either 0 because its frictionless or 73N to equal out the 73N of the sled as well and preventing it from sinking thru the hill?
 
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this whole thing with the angle is what throws me off on this problem
 
anglum said:
the force on the rope would be 73N to equal out the 73N of the sled

and the force the hill exerts on the sled would be either 0 because its frictionless or 73N to equal out the 73N of the sled as well and preventing it from sinking thru the hill?

No, don't forget the normal force the hill exerts.

draw a freebody diagram... what are the forces acting along the incline?
 
woudlnt the force on the rope just be 73N to equual that of the sled tho?
 
anglum said:
woudlnt the force on the rope just be 73N to equual that of the sled tho?

No. Draw the freebody diagram of the sled. There is a normal force that acts perpendicular to the incline... Gravity acts downwards... divide gravity into two components... one that acts perpendicular to the incline(opposite the direction of the normal)... one that acts parallel to the incline...

What is the component of gravity perpendicular to the incline... what is the component of gravity parallel to the incline.
 
do u mean draw it out liek a triangle and calculate the forces straight down and also the forces to the right?
 
then use sin34 to find out the down force and the horizontal force

and get down force to be 40.82 and horizontal force to be 60.519?
 
or am i not even close to doing this the right way?
 
  • #10
anglum said:
or am i not even close to doing this the right way?

You are very close... make sure you have the angles right. What is the angle between the the gravity vector downwards... and the component perpendicular to the incline?
 
  • #11
the angles are 90 degrees 34 degrees and 56 degrees?

i cannot figure out what formulas to use to figure this out... this problem is confusing when it shouldn't be blahhh

keep in mind i need the force exerted on the rope

and the force exerted on the sled by the hill
 
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  • #12
anglum said:
the angles are 90 degrees 34 degrees and 56 degrees?

i cannot figure out what formulas to use to figure this out... this problem is confusing when it shouldn't be blahhh

Yes, but which one is 34 and which one is 56? Look at the triangle formed by gravity and its two components... you need an angle in that triangle... then you can use sin and cos...
 
  • #13
the angle with gravity is 56 degrees... but the problem says nothing about gravity so why am i takin that into accoutn?

i need to know that for the force exerted on the rope?
 
  • #14
anglum said:
the angle with gravity is 56 degrees... but the problem says nothing about gravity so why am i takin that into accoutn?

i need to know that for the force exerted on the rope?

you can't get the tension in the rope until you know the component of gravity acting along the incline...

Look at this page: under where it says "frames of reference":

http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/Dynamics/InclinePlanePhys.html

see the two components of gravity?
 
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  • #15
ok i see the 2 components of gravity... and am i right in saying the 3 angles in the problem are 34, 56, and 90? those are the 3 i need to solve this??
 
  • #16
anglum said:
ok i see the 2 components of gravity... and am i right in saying the 3 angles in the problem are 34, 56, and 90? those are the 3 i need to solve this??

yes. 34 is what you need. it's the [tex]\theta[/tex] angle like in the link I sent. So what is the component of gravity perpendicular to the plane? What is the component of gravity parallel to the plane?
 
  • #17
so its mass*gravity sin34

and the one perpendicular is mass *gravity cos 34?
 
  • #18
anglum said:
so its mass*gravity sin34

and the one perpendicular is mass *gravity cos 34?

yes, so in this case... the parallel component is 73*sin34. And the perpendicular component is 73*cos34.

so taking the sled and rope as one system... what are the two forces acting on this system parallel to the plane? This will give you the answer to the first part...

what are the forces acting perpendicular to the plane? That gives the secod part.
 
  • #19
well the parallel component is equal to 40.821
the perp is 60.519
the 2 forces acting on this system are 73N and 40.821N?
 
  • #20
anglum said:
well the parallel component is equal to 40.821
the perp is 60.519
the 2 forces acting on this system are 73N and 40.821N?

how are you getting these numbers?

oops... never mind I had my calculator wrongly setup...

ok... so the two forces parallel are 40.821 and the force the boy exerts on the rope... so what's the force the boy exerts on the rope?
 
  • #21
parallel was 73*sin34 and perp was 73*cos34

and then the 73N was given in the problem
 
  • #22
there is a diagram in the problem where the sled is labeled as a box with 73N written in it
 
  • #23
anglum said:
parallel was 73*sin34 and perp was 73*cos34

and then the 73N was given in the problem

yeah, sorry. had my calculator setup in radians.
 
  • #24
hahah ok... i made that mistake the other day...

so i have the parallel as 40.821N is that the answer for the first part?
 
  • #25
anglum said:
hahah ok... i made that mistake the other day...

so i have the parallel as 40.821N is that the answer for the first part?

yup exactly. same type of thing for the second part, except using the perpendicular.
 
  • #26
so the force on the rope is 40.821 and the force the hill exerts on the sled is 60.519 ?
 
  • #27
anglum said:
so the force on the rope is 40.821 and the force the hill exerts on the sled is 60.519 ?

yup.
 
  • #28
sweet.. i am going to create a new topic for the next problem ok
 

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