Simple Physics Mass pulled by a Force

In summary, the problem involves calculating whether a mass can be lifted off the ground and whether it will move along the ground when pulled by a force. Part 1 of the problem shows that the mass cannot be lifted off the ground. Part 2 involves calculating the maximum values of static and kinetic friction and comparing them to the horizontal force being applied. The result is that the mass will move along the ground.
  • #1
c-murda
67
0
[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?
 

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  • #2
c-murda said:
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?
 
  • #3
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?


Doc Al said:
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!
 
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  • #4
c-murda said:
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.
 
  • #5
Doc Al said:
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?
 
  • #6
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.
 
  • #7
Doc Al said:
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?
 
  • #8
Almost. (Check your signs.)
 
  • #9
Doc Al said:
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?...
 
  • #10
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?
 
  • #11
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
 
  • #12
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.
 
  • #13
Doc Al said:
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?
 
  • #14
c-murda said:
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.
 
  • #15
Doc Al said:
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
 
  • #16
That's what I mean. :wink:
 
  • #17
which could only mean...

Fs = 11.9 and Fk = 6.36
are these Newtons?


...then do i compare to f cos theta?
 
  • #18
Yes.
 
  • #19
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 = Vot + (.5)(a)(t^2)...

i think...and we must solve for acceleration first?
 
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  • #20
c-murda said:
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 = Vot + (.5)(a)(t^2)...
Good. In this case, we presume the box is initially at rest.
 
  • #21
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
 
  • #22
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
 
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  • #23
c-murda said:
A+...97 is good enough for me
Outstanding! :approve:
 
  • #24
is there a way to mark this page as SOLVED CORRECTLY?
 
  • #25
Absolutely: https://www.physicsforums.com/showpost.php?p=1501730&postcount=3"
 
Last edited by a moderator:

Related to Simple Physics Mass pulled by a Force

1. What is the definition of "Simple Physics Mass pulled by a Force"?

Simple Physics Mass pulled by a Force is a basic concept in physics that describes the relationship between mass and force. It states that when an object with a certain mass is subjected to a force, it will accelerate in the direction of the force.

2. How is mass related to the amount of force required to move an object?

According to the principle of simple physics, the greater the mass of an object, the more force is needed to move it. This means that a heavier object will require more force to accelerate or move than a lighter object.

3. What is the formula for calculating force in simple physics?

The formula for calculating force in simple physics is F = m x a, where F is force, m is mass, and a is acceleration. This means that force is directly proportional to mass and acceleration.

4. Can an object with a smaller mass exert a greater force than an object with a larger mass?

No, an object with a smaller mass cannot exert a greater force than an object with a larger mass. This is because the amount of force an object can exert is directly proportional to its mass. So, the greater the mass, the greater the force it can exert.

5. How does the direction of the force affect the motion of an object?

In simple physics, the direction of the force is important in determining the motion of an object. If the force is applied in the same direction as the object's motion, it will increase its speed. If the force is applied in the opposite direction, it will decrease its speed. And if the force is applied perpendicular to the direction of motion, it will change the direction of motion without changing the speed.

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