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

[Muteb]

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

 

[PhanthomJay]

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.

 

[Muteb]

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

 

[PhanthomJay]

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.

 

[Muteb]

a=((1.75^2) +(-7.06^2))^1/2 =7.28m/s^2

how about it now

 

[Muteb]

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

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a=((1.75^2) +(-2.35^2))^1/2 =2.93m/s^2

 

[PhanthomJay]

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

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a=((1.75^2) +(-2.35^2))^1/2 =2.93m/s^2

Now you're catching on! You are exactly right.