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?
7.37 m/s^2
What is the magnitude of the normal force exerted on the mass by the floor?
3.68×101 N
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?
NEW
Anonymous 2 (Wed Jan 20 06:34:59 pm 2010 (EST))
did anyone figure out how to do the second part?
NEW
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
NEW
What formula are we using for part 1?
NEW
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 :(
NEW
For the first one you use mass(a)= (force given)*cos theta,
second one is mass*9.8-force*(sin theta) hope it helps
NEW
Anonymous 6 (Sat Jan 23 08:49:14 pm 2010 (EST))
what are the units for each???
NEW
The first is m/s^2 and the second is N.
NEW
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.
NEW
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?
NEW
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
NEW
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?
NEW
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??
NEW
Re: Anonymous 11 (Sun Jan 24 04:41:58 pm 2010 (EST))
thanks stephanie
NEW
Someone should have probably mentioned that your calculator
has to be on DEGREE and not RADIANS! I hate trig...
NEW
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.
is the normal force equal to the y component of the vector?
or it it just the weight?
NEW
Anonymous 2 (Wed Jan 20 06:34:59 pm 2010 (EST))
did anyone figure out how to do the second part?
NEW
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
NEW
What formula are we using for part 1?
NEW
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 :(
NEW
For the first one you use mass(a)= (force given)*cos theta,
second one is mass*9.8-force*(sin theta) hope it helps
NEW
Anonymous 6 (Sat Jan 23 08:49:14 pm 2010 (EST))
what are the units for each???
NEW
The first is m/s^2 and the second is N.
NEW
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.
NEW
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?
NEW
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
NEW
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?
NEW
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??
NEW
Re: Anonymous 11 (Sun Jan 24 04:41:58 pm 2010 (EST))
thanks stephanie
NEW
Someone should have probably mentioned that your calculator
has to be on DEGREE and not RADIANS! I hate trig...
NEW
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.