Simple Physics Mass pulled by a Force

[c-murda]

[SOLVED] Simple Physics Mass pulled by a Force

Homework Statement

Two part problem:
Part 1:

Show by calculations that the mass can not be lifted up from the ground

F=27.0
m=3.00kg
Angle= 30degrees

Part 2:
show by calculations whether the mass moves level along the ground, if the coefficient of static friction between the surface and the mass is .750 and the coefficient of kinetic friction is .400

attached is drawing...

Homework Equations

The Attempt at a Solution

i draw the FBD and get everything in terms of x an y

Fx=F cos 30 = m*a; 27.0 cos 30 = m*a; a = 7.79 m/s^2
Fy= Fn sin 30 - m*g = 0 ;Fn = (m*g)/(sin 30) , which equaled 58.8N

uhmm... or do i

w = m*g; 3.00*9.80 = 29.4
f sin 30 = 13.5

13.5> 29.4 so, obj doesn't move up or down?

im confused...

Part 2:

F=27.0
m= 3.00kg
Fs=.750
Fk=.400

I draw FBD and

F cos 30 = 23.4

Fx= F cos 30 - Fk = m*a; a = (27.0 cos 30 -.400)/3 = 7.66m/s^2

so the object moves right?

 

[Doc Al]

i draw the FBD and get everything in terms of x an y

Fx=F cos 30 = m*a; 27.0 cos 30 = m*a; a = 7.79 m/s^2
Fy= Fn sin 30 - m*g = 0 ;Fn = (m*g)/(sin 30) , which equaled 58.8N

uhmm... or do i

w = m*g; 3.00*9.80 = 29.4
f sin 30 = 13.5

13.5> 29.4 so, obj doesn't move up or down?

im confused...

Just compare the upward component of the force with the weight. (Which is what you did in your second attempt.)

For the second part, realize that

Part 2:

F=27.0
m= 3.00kg
Fs=.750
Fk=.400

Fs & Fk are the coefficients of friction, not the force of friction. Hint: What's the normal force?

 

[c-murda]

so since the weight is more than the upward force(f sin 30) there is no upward movement?

does a normal force come into play here? is it equal to m*g?

Fs & Fk are the coefficients of friction, not the force of friction. Hint: What's the normal force?

i know they are the coefficients of friction but, i didn't know how to type uhmm... Mu? (spelling)...do you know how to type that I am noob so i don't...

thnx doc!

 

[Doc Al]

so since the weight is more than the upward force(f sin 30) there is no upward movement?

Right. In order to lift the mass, the force must support its weight. But it weighs too much for that.

does a normal force come into play here? is it equal to m*g?

The normal force does not equal mg. (It would, if there were no applied force.) To find the normal force, add up all the vertical force components. What must the sum equal?

i know they are the coefficients of friction but, i didn't know how to type uhmm... Mu?

I'm not concerned with how you write it, but that you don't leave out the normal force when calculating friction.

 

[c-murda]

The normal force does not equal mg. (It would, if there were no applied force.) To find the normal force, add up all the vertical force components. What must the sum equal?

well the question ask if the box moves right along the level of the ground...with the given coefficents of friction. so, i need to figure out if the force pulling the box is greater than the force restricting the box(normal force?) correct?

 

[Doc Al]

You need to compare the horizontal component of the applied force with the force of friction. Start by finding the normal force, which you'll need in order to calculate the friction force.

 

[c-murda]

You need to compare the horizontal component of the applied force with the force of friction. Start by finding the normal force, which you'll need in order to calculate the friction force.

okay how about,

Fn = F sin (theta) - m*g?

 

[Doc Al]

Almost. (Check your signs.)

 

[c-murda]

Almost. (Check your signs.)

lol...F sin (theta) + m*g?

i would assume that would be correct...then once i have the normal force where would i go from there?...

 

[Doc Al]

That's not correct. The way to do it is to add up the vertical force components (each with the correct sign) and set the sum equal to zero. Then you can solve for Fn.

How does friction relate to the normal force and mu?

 

[c-murda]

im sorry this problem has me stumped...

the only vertical force would be...f sin theta and the normal force... opposite force would be m*g

horizontal forces would be friction to the left...and f cos theta?

would it be Fn = m*g - F sin theta

 

[Doc Al]

Good.

Now compare F cos theta to the friction force. First figure out the maximum value of static friction and check if that horizontal force is enough to overcome it.

 

[c-murda]

Good.

Now compare F cos theta to the friction force. First figure out the maximum value of static friction and check if that horizontal force is enough to overcome it.

is Fn = 15.9?


so f cos theta = 27.0 cos 30 = 23.4

so... Fx = F1 cos 30 - f(mu) = ma?

then solve for accel?

 

[Doc Al]

is Fn = 15.9?

Good.

so f cos theta = 27.0 cos 30 = 23.4

Good.

so... Fx = F1 cos 30 - f(mu) = ma?

then solve for accel?

First figure out the friction forces.

 

[c-murda]

Good.


Good.


First figure out the friction forces.

sorry it has been along time since physics...uhmm...by that you mean... like


Fs= Fn*(mu)s

and

Fk= Fn*(mu)k

?...lol

 

[Doc Al]

That's what I mean.

 

[c-murda]

which could only mean...

Fs = 11.9 and Fk = 6.36
are these Newtons?


...then do i compare to f cos theta?

 

[Doc Al]

Yes.

 

[c-murda]

would the sum of all x forces be? Fx = F cos theta - (mu)k = m*a

since f cos theta > Fs and f cos theta > Fk

since those are magnitudes of the friction forces that means the box will move...

and its displacement is given by? (not part of the problem...just curious)
y = volt + (.5)(a)(t^2)...

i think...and we must solve for acceleration first?

 

[Doc Al]

would the sum of all x forces be? Fx = F cos theta - (mu)k = m*a

since f cos theta > Fs and f cos theta > Fk

since those are magnitudes of the friction forces that means the box will move...

Right. Since the applied force is greater than static friction, the box will move.

To find the acceleration of the box, just find the net force and apply Newton's 2nd law.

and its displacement is given by? (not part of the problem...just curious)
y = volt + (.5)(a)(t^2)...

Good. In this case, we presume the box is initially at rest.

 

[c-murda]

thanks doc your help was much appreciated i will let you know what i get on my test...lol...the third question was posted if your interested in looking at it...

c

 

[c-murda]

so my teacher took off a couple points because i didn't say that it would move horizontally...when that is what the question was asking. i simply said yes the box will move, assuming he was asking if it was going horizontally and he also asked why?

A+...97 is good enough for me


c-murda

 

[Doc Al]

A+...97 is good enough for me

Outstanding!

 

[c-murda]

is there a way to mark this page as SOLVED CORRECTLY?

 

[Doc Al]

Absolutely: https://www.physicsforums.com/showpost.php?p=1501730&postcount=3"