Calculating Magnitude of Acceleration for 3.0 kg Mass w/ 2 Forces

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Homework Help Overview

The discussion revolves around calculating the magnitude of acceleration for a 3.0 kg mass subjected to two forces: one of 9.0 N directed east and another of 8.0 N directed at an angle of 62.0 degrees north of west. The problem specifies that there is no gravitational force acting on the mass.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the calculation of acceleration using the formula Fx=ma, breaking down the forces into their components. There are attempts to calculate both the x and y components of acceleration, with some questioning the accuracy of previous calculations and suggesting the use of Pythagorean theorem to find the resultant acceleration.

Discussion Status

Several calculations have been presented, with participants providing feedback on each other's math. There is an ongoing exploration of the correct approach to derive the final magnitude of acceleration, with some participants offering corrections and adjustments to previous values. The discussion reflects a collaborative effort to clarify the calculations involved.

Contextual Notes

Participants note the absence of gravitational forces and emphasize the need to divide forces by mass to find acceleration components. There is a focus on ensuring that all calculations adhere to the principles of physics without overlooking any necessary steps.

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Only two foces act on a 3.0 kg mass.?
Only two foces act on a 3.0 kg mass. One of the forces is 9.0 N east, and the other is 8.0 N in the direction of 62.0 north of wet. What i the magnitude of the acceleration of the mass? The mass has only TWO forces acting on it. No gravity. Not resting on any other surface

Fx=ma

9cos0-8cos62=3a
a=1.7m/^2

i that rigth
 
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Muteb said:
Only two foces act on a 3.0 kg mass.?
Only two foces act on a 3.0 kg mass. One of the forces is 9.0 N east, and the other is 8.0 N in the direction of 62.0 north of wet. What i the magnitude of the acceleration of the mass? The mass has only TWO forces acting on it. No gravity. Not resting on any other surface

Fx=ma

9cos0-8cos62=3a
a=1.7m/^2

i that rigth
That's the acceleration in the east or positive x direction. You need also to calculate the acceleration in the northerly or positive y direction, then find the magnitude of the acceleration using Pythagorus' Theorem.
 
Fx=ma
9cos0-8cos62=3a
a=1.7m/^2
Fy=ma
9sin0-8sin62=ma
ay= 1.97
a=(1.2^2 + 1.97^2)^1/2 a=2.30 m/s^2 is it right now and thanks for your help
 
Muteb said:
Fx=ma
9cos0-8cos62=3a
a=1.7m/^2
Fy=ma
9sin0-8sin62=ma
ay= 1.97
a=(1.2^2 + 1.97^2)^1/2 a=2.30 m/s^2 is it right now and thanks for your help
Well, your math is not very good. 9cos0 = 9, and 8cos62 = 3.75. ax = 1.75m/sec^2.

9sin0 is 0, and 8sin62 is 7.06. ay = -7.06m/sec^2. Now solve for a.
 
a=((1.75^2) +(-7.06^2))^1/2 =7.28m/s^2

how about it now
 
but you forget to divide the Fy by Mass that -7.06/3 to get the ay that is -2.35

--------------------------------------------------------------------------------

a=((1.75^2) +(-2.35^2))^1/2 =2.93m/s^2
 
Muteb said:
but you forget to divide the Fy by Mass that -7.06/3 to get the ay that is -2.35

--------------------------------------------------------------------------------

a=((1.75^2) +(-2.35^2))^1/2 =2.93m/s^2
Now you're catching on! You are exactly right.
 

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