## Simple constant force problem

A block of mass 6.5 kg is pulled along a horizontal frictionless floor by a cord that exerts a force of 55.0 N at an angle 29.4° above the horizontal.

What is the magnitude of the acceleration of the block?

What is the magnitude of the normal force exerted on the mass by the floor?

What is the magnitude of the acceleration of the block?

**7.37 m/s^2**What is the magnitude of the normal force exerted on the mass by the floor?

**3.68×101 N**## Discussion

*Anonymous 2*(Wed Jan 20 01:12:32 pm 2010 (EST))

is the normal force equal to the y component of the vector?

or it it just the weight?

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*Anonymous 2*(Wed Jan 20 06:34:59 pm 2010 (EST))

did anyone figure out how to do the second part?

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check it

for the second part draw a diagram and lable all forces

acting on the y axis.

on y+: Normal force and y component of the f vector.

on y-: weight aka m*g

all forces on the y axis must cancel out right so...

0=N+Fy-m*g

solve for N

hope this helps

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What formula are we using for part 1?

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How do we do part one. I've done it a few different ways,

and all of them are wrong :( I have one try left :(

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For the first one you use mass(a)= (force given)*cos theta,

second one is mass*9.8-force*(sin theta) hope it helps

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*Anonymous 6*(Sat Jan 23 08:49:14 pm 2010 (EST))

what are the units for each???

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The first is m/s^2 and the second is N.

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*Anonymous 6*(Sat Jan 23 10:17:09 pm 2010 (EST))

Ok I dont know what Im doing wrong. For the first one I use

the eqn m = (force)(sin theta) so I have (40)(cos25.6) =

36.07 m/s^2 Idk what im doing wrong.

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*Anonymous 6*(Sat Jan 23 10:19:22 pm 2010 (EST))

For the second one I have (mass)(9.8-force)(sin theta). So

I have (6.7)(9.8-40)(sin 25.6) I get -87.4. I tried to

change it to a positive and this is wrong as well. Help

anyone?

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The force F has two components ,

Horizontal component = F cos θ=53* cos 27.9o = 46.84 N

Vertical component = F sin θ=53* sin 27.9o = 24.8 N

Also there is no other force in horizontal direction ,

So acceleration = horizontal force /mass = 46.84 / 7.5 =

6.245 m/s2

Here forces in vertical direction are

1 . Gravitational force =mg downwards =7.5*9.8=73.5 N

2 . Normal force =N upwards

3. Vertical component = F sin θ=53* sin 27.9o = 24.8 N

(upwards )

Also there will be no acceleration in vertical direction ,

SO N +24.8 = 73.5

N = 48.7 N

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I have tried 0=N+Fy-m*g, but that has not helped solving for N...

i tried...0=N+54-5.47*9.8...getting the wrong answer...anyone have a better

formula for me to try for part 2?

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with theta=31.5 and force=44N I used the equation

a=horizontalforce/mass which was a= F*cos(31.5)/6.9kg which

calculated to a= 43.845/6.9kg = 6.35m/s^2. I also

calculated this number following example 4.1 on page 93 in

the book (excluding the friction factor) and came up with

a=6.35m/s^2 again. Why does Lon Capa say this is incorrect??

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**Re:**

*Anonymous 11*(Sun Jan 24 04:41:58 pm 2010 (EST))

thanks stephanie

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Someone should have probably mentioned that your calculator

has to be on DEGREE and not RADIANS! I hate trig...

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I'm on my last try can someone give me an equation to use on part 2? I'm stuck

and I need help bad.