- #1

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wut i did is:

F sin 37-0.1+mg=0

Fcos37=n

but i didnt get the answer right which is 63.2 !!!

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- Thread starter mooneh
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- #1

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wut i did is:

F sin 37-0.1+mg=0

Fcos37=n

but i didnt get the answer right which is 63.2 !!!

- #2

Doc Al

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Redo this correctly: The sum of forces in the vertical direction must add to zero.F sin 37-0.1+mg=0

Good.Fcos37=n

Is the given angle

- #3

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but even when i add them up the answer is wrong

and by solving Fcos37=n the answer of F is wrong too !!

- #4

Doc Al

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What are the vertical forces? Which way do they act?

- #5

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the vertical forces acts upward because its abt to slip down

right ?

- #6

HallsofIvy

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- #7

Doc Al

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Gravity acts upward? What's 0.1 supposed to represent? What's the friction

the vertical forces acts upward because its abt to slip down

right ?

Yep. But that's not the only problem.

- #8

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here is wut i did:

Fsin37+0.2-mg=0

Fcos 37 =n

F ( cos37+sin37) = 93.88

F = 67

Fsin37+0.2-mg=0

Fcos 37 =n

F ( cos37+sin37) = 93.88

F = 67

- #9

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Well you fixed one thing, broke another, and are missing a big thing in both cases >_>

There are three forces acting on it, friction, the guy pushing, and gravity. Gravity and friction are both trying to pull it down, the guy pushing is trying to push it up. Since they're in the same direction, gravity and friction need to be both negative or both positive, and then the guy pushing should be the opposite

As of your previous post you have...actually I'm looking at your work and maybe you did it more right than it seemed

You typed Fsin(37)+.2-mg=0, and I'm looking at the .2, .2 by itself is not the force of friction, you must multiply by the normal force. But then you found the normal force and used it but I dunno where the .2 went or the 93.88 came from. You have the right idea though

Remember, force of friction = coefficient * normal force

Fgrav+Ffr-F*sin37=0, or F*sin37-Fgrav-Ffr=0, whichever

There are three forces acting on it, friction, the guy pushing, and gravity. Gravity and friction are both trying to pull it down, the guy pushing is trying to push it up. Since they're in the same direction, gravity and friction need to be both negative or both positive, and then the guy pushing should be the opposite

As of your previous post you have...actually I'm looking at your work and maybe you did it more right than it seemed

You typed Fsin(37)+.2-mg=0, and I'm looking at the .2, .2 by itself is not the force of friction, you must multiply by the normal force. But then you found the normal force and used it but I dunno where the .2 went or the 93.88 came from. You have the right idea though

Remember, force of friction = coefficient * normal force

Fgrav+Ffr-F*sin37=0, or F*sin37-Fgrav-Ffr=0, whichever

Last edited:

- #10

Doc Al

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0.2 is just thehere is wut i did:

Fsin37+0.2-mg=0

- #11

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Fsin37-0.2n-n=0

Fsin37= 1.2(Fcos37)

???

Fsin37= 1.2(Fcos37)

???

- #12

Doc Al

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Which way does the friction force act? What happened to the weight?Fsin37-0.2n-n=0

- #13

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thats why i changed my answer and put negative signs to both Ffr and weight

weight=mg=n

- #14

Doc Al

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But since the block is just about to slideblochwave said Fgrav+Ffr-F*sin37=0, or F*sin37-Fgrav-Ffr=0

thats why i changed my answer and put negative signs to both Ffr and weight

Weight = mg, but doesweight=mg=n

The above is just the equation for vertical forces. Combine this (after you fix it) with your equation for the horizontal forces (from your first post).

- #15

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Fcos37=n

Fsin37+ 0.2n - mg=0

5 Fsin37 + n -235.2 = 0

Fcos 37 - n =0

F(5sin37+cos37)=235.2

F=61.7

Fsin37+ 0.2n - mg=0

5 Fsin37 + n -235.2 = 0

Fcos 37 - n =0

F(5sin37+cos37)=235.2

F=61.7

- #16

Doc Al

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Looks good to me!

- #17

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never mind close enough :)

lool u must hate me now

thx alooooooooot

ur the best

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