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Dynamics -- Newton's 2nd law with acceleration

  • Thread starter Nanu Nana
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  • #1
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Homework Statement


A sleigh with a mass of 150 kg from rest is horizontally towed to the left
. This is done using a rope that makes an angle of 30 ° with the horizontale. The tensile force is 200 N. (Neglect friction forces.)
a. What is the horizontal component of the force?
b. What acceleration will the sleigh get ?

Homework Equations


F: m xa[/B]


The Attempt at a Solution


for a ) horizontal component zo
Fx = 200 N x cos (30°) = 173.2N
For b)
a= F/m => 173.2N / 150 kg = 1.155m/s^2
For b ) i thought we had to use tensile force ?? why do we use Fx instead of tensile force ???[/B]
 

Answers and Replies

  • #2
BvU
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Well, would Fx = 0 (pulling vertically) result in horizontal acceleration ?
 
  • #3
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Well, would Fx = 0 (pulling vertically) result in horizontal acceleration ?
SOrry I don't understand??
 
  • #4
BvU
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What acceleration do you expect in case Fx = 0 ?

upload_2016-8-23_17-48-2.png
 
  • #5
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200/150 = 1.33 m/s^2
 
  • #6
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Earth is pulling down with 150 kg x 9.81 m/s2 = 1470 N. Who wins ?
 
  • #7
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Earth is pulling down with 150 kg x 9.81 m/s2 = 1470 N. Who wins ?
Gravity wins
 
  • #8
BvU
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Yes. So the vertical component of the tension is offset by gravity (*). Only the horizontal component can cause horizontal acceleration.

(*) The other force that acts is the normal force from the ground, that keeps the sleigh from disappearing into the earth. So unless the vertical componenet of the tension is > 1470 N, gravity wins and there is no vertical acceleration.
 
  • #9
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The acceleration of vertical component of the tension and acceleration of tension is same ???
 
  • #10
BvU
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I don't know what you mean with acceleration of tension ?
Unless you are referring to the picture, where I drew the pulling force in a vertical direction. Then the tension and the vertical component of the tension are the same.
In your exercise the pulling force is at 30 degrees.
 
  • #11
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For b ) I thought we had to use tensile force ?? Why do we use Fx instead of tensile force ???
Break down the 200 N tensile force into two perpendicular components (Fx and Fy).
Fx + Fy = 200 N pointing 30 degrees above horizontal.
(Note we use vector addition in this equation.)
Keeping in mind that force and acceleration are parallel, select the portion of the tensile force that is relevant to your problem, and ignore the component that is irrelevant.
 
Last edited:

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