- #1

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a) how large is the frictoinal force?

b) what is the coefficient of friction?

For Ans A, how would i find frictional force... 40N-5x6 = 10 N??

thanks

- Thread starter jai6638
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- #1

- 263

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a) how large is the frictoinal force?

b) what is the coefficient of friction?

For Ans A, how would i find frictional force... 40N-5x6 = 10 N??

thanks

- #2

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i cant seem to use MA=Fapp - Ff

i.e. ma = fpp - ukmg sice i dont have the value of a...

- #3

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set up a FBD for the block

Use Newton's second Law for both horizontal and vertical directions

[tex]\vec{F} = m\vec{a} [/tex]

They give the acceleration in the horizontal direction... what is the accel. in the vertical direction??

What are all the forces acting on the block?

The equation for the force of friction is

[tex]\vec{F}_f = \mu_kN[/tex]

where [itex]\mu_k[/itex] is the coefficient of friction and [itex]N[/itex] is the Normal force

your answer for part A is correct

show your work for more help

Use Newton's second Law for both horizontal and vertical directions

[tex]\vec{F} = m\vec{a} [/tex]

They give the acceleration in the horizontal direction... what is the accel. in the vertical direction??

What are all the forces acting on the block?

The equation for the force of friction is

[tex]\vec{F}_f = \mu_kN[/tex]

where [itex]\mu_k[/itex] is the coefficient of friction and [itex]N[/itex] is the Normal force

your answer for part A is correct

show your work for more help

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- #4

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The definition of acceleration isjai6638 said:

i cant seem to use MA=Fapp - Ff

i.e. ma = fpp - ukmg sice i dont have the value of a...

[tex]\vec{a} = \frac{d\vec{v}}{dt} [/tex]

ie: the change in velocity wrt time.

here they give you that the velocity is constant so that should tell you what the acceleration is.

- #5

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F= MA

F/M=a

therefore , A= 30/12. = 2.5 m/s^2

am i correct?

- #6

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- #7

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Not quite... I gave you a big hint in the previous postjai6638 said:

F= MA

F/M=a

therefore , A= 30/12. = 2.5 m/s^2

am i correct?

-for one you need to draw a Free Body Diagram (FBD) for the crate

- identify All the forces acting on the crate

-Then you need to break newton's second law into vertical and horizontal components... do this and show your work (list the forces along with their directions dont try to draw the FBD) and I'll help you further.

as for the acceleration.. you need to realize that newtons 2nd law equates the NET force (the sum of all forces) ...you forgot about one

- since the velocity is CONSTANT that means that its not changing wrt time in the horizontal direction so the acceleration would be 0 right?

- #8

futb0l

It's is at a CONSTANT VELOCITY. Therefore the acceleration is = ...jai6638 said:

F= MA

F/M=a

therefore , A= 30/12. = 2.5 m/s^2

am i correct?

- #9

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ALso, just for my information, if there is constant velocity is Ffrictoin = F applied?

- #10

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Don't forget about the friction force... what direction does it point in?jai6638 said:

ALso, just for my information, if there is constant velocity is Ffrictoin = F applied?

In this case yes since

[tex]\Sigma\vec{F}_x = \vec{F}_h +\vec{F}_f[/tex]

[tex]m\vec{a} = 12*0 [/tex]

[tex]\Rightarrow \vec{F}_h +\vec{F}_f = 0[/tex]

[tex]\Rightarrow \vec{F}_h = -\vec{F}_f [/tex]

[tex]\Rightarrow F_f = -F_h [/tex]

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- #11

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Frictoinal force points to the left...

- #12

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I gave you an example of using newtons 2nd law for the horizontal components ... why don't you try it now for the vertical components...

keep in mind what your trying to solve for

keep in mind what your trying to solve for

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- #13

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im gonna sleep since i am not accomplishing much by stayin awake as all the answers/equations are goin above my head .. makes me wonder if im fit for becoming an engineer... doh anyways... thanks for ur help though.. appreciate it :)

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