Dynamics -- Newton's 2nd law with acceleration

AI Thread Summary
The discussion focuses on solving a physics problem involving a sleigh being towed with a force at an angle. The horizontal component of the force is calculated as 173.2 N using the cosine of the angle. The resulting acceleration of the sleigh is determined to be 1.155 m/s² by dividing the horizontal force by the mass of the sleigh. Clarification is provided that only the horizontal component of the tensile force contributes to horizontal acceleration, while the vertical component is counteracted by gravity. Understanding the breakdown of the tensile force into its components is crucial for solving the problem accurately.
Nanu Nana
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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]
 
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Well, would Fx = 0 (pulling vertically) result in horizontal acceleration ?
 
BvU said:
Well, would Fx = 0 (pulling vertically) result in horizontal acceleration ?
SOrry I don't understand??
 
What acceleration do you expect in case Fx = 0 ?

upload_2016-8-23_17-48-2.png
 
200/150 = 1.33 m/s^2
 
Earth is pulling down with 150 kg x 9.81 m/s2 = 1470 N. Who wins ?
 
BvU said:
Earth is pulling down with 150 kg x 9.81 m/s2 = 1470 N. Who wins ?
Gravity wins
 
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.
 
The acceleration of vertical component of the tension and acceleration of tension is same ?
 
  • #10
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
Nanu Nana said:
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.
 
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