Forces on an Object: Find Acceleration, Mass, Speed & Velocity Components

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Homework Help Overview

The discussion revolves around a physics problem involving forces acting on an object, where participants are tasked with determining the direction of acceleration, mass, speed after a certain time, and velocity components. The subject area includes dynamics and vector analysis.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between force and acceleration, questioning how to determine the direction of the acceleration vector using Newton's second law. There are attempts to calculate angles and resultant forces, with some participants expressing uncertainty about their initial drawings and calculations.

Discussion Status

The discussion is ongoing, with participants providing guidance on understanding the direction of acceleration and its relationship to velocity components. There is recognition of correct calculations, but some participants are still grappling with the connections between the various components of the problem.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information they can use or the methods they can apply. There is an emphasis on understanding vector directions and relationships without providing direct solutions.

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Homework Statement


Three forces acting on an object are given by F1 = ( 2.20 j ) N, F2 = ( 3.45 j ) N, and F3 = ( - 40.0 i ) N. The object experiences an acceleration of magnitude 3.95 m/s2.

(a) What is the direction of the acceleration?

(b) What is the mass of the object?

(c) If the object is initially at rest, what is its speed after 13.0 s?

(d) What are the velocity components of the object after 13.0 s?

Homework Equations


f=ma pythagorean theorem

The Attempt at a Solution



(a)I drew this triangle to find both the resultant force and angle.
http://img255.imageshack.us/img255/2737/trianglejb6.th.jpg

I calculated theta to be 171.9602 deg from positive x-axis using tan theta=(40/5.65)
and calculated R to be 40.3971N.

(b)Taking R from (a), I have f=ma --> m=f/a m=40.3971/3.95=10.2271kg

(c) Using vf=v0t+at we have vf=(0)(13)+(3.95)(13)=51.35m/s

(d)This is the one I'm lost on. I have no idea how to convert to velocity vector.Anyone have any insight on (d)? Also, does the rest of the work look to be correct? Thanks in advance.
 
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You should first determine the direction of the acceleration vector in part (a), using Newton's 2nd law. That should help to understand the direction of the velocity vector. The rest looks OK. You're almost there.
 
PhanthomJay said:
You should first determine the direction of the acceleration vector in part (a), using Newton's 2nd law. That should help to understand the direction of the velocity vector. The rest looks OK. You're almost there.

How would I relate Newton's second law to the direction of the acceleration vector? Isn't it just f=ma?
 
sailsinthesun said:
How would I relate Newton's second law to the direction of the acceleration vector? Isn't it just f=ma?
Yes, it's f_net = ma, where f and a are vector quantities. If f acts in a certain direction, in what direction must a act?
 
PhanthomJay said:
Yes, it's f_net = ma, where f and a are vector quantities. If f acts in a certain direction, in what direction must a act?

The same direction as the force vector, correct?

I still don't think I'm understanding. I take it I drew the triangle wrong initially? I took the three forces to be (2.2N@90deg), (3.45N@90deg), and (-40N@180deg), so theta in the triangle is 81.9602 deg which is 171.9602deg from the positive x-axis.
 
Last edited:
sailsinthesun said:
The same direction as the force vector, correct?

I still don't think I'm understanding. I take it I drew the triangle wrong initially? I took the three forces to be (2.2N@90deg), (3.45N@90deg), and (-40N@180deg), so theta in the triangle is 81.9602 deg which is 171.9602deg from the positive x-axis.
Why do you think you drew it incorrectly? The resultant force acts 171.9 degrees from the +x axis, this is correct. So what's the direction of the acceleration with respect to the +x axis?
 
PhanthomJay said:
Why do you think you drew it incorrectly? The resultant force acts 171.9 degrees from the +x axis, this is correct. So what's the direction of the acceleration with respect to the +x axis?

Also 171.9 degrees. I guess I'm struggling to relate this angle with the velocity components asked for in part (d).
 
sailsinthesun said:
Also 171.9 degrees. I guess I'm struggling to relate this angle with the velocity components asked for in part (d).
Yes, and since the object starts from rest, v=at (as you noted), and since a and v are vectors, and you now know the direction of a, what's the direction of v? Then break up v, which you have already calculated, into its x and y components using basic trig, or by comparison to the force components using similar triangles.
 

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