# Force when dragging a suitcase at constant velocity

## Homework Statement

You drag a suitcase of mass 17 kg with a
force of F at an angle 40.6◦ with respect to
the horizontal along a surface with kinetic
coefficient of friction 0.59.
The acceleration of gravity is 9.8 m/s2 .
If the suitcase is moving with constant ve-
locity 2.68 m/s, what is F?

Ff = miuFn
All forces equal

## The Attempt at a Solution

so first i broke it into components,
Ff = Fcostheta
Fn + Fsin theta = Fg
Ff = miuFn = miu( Fg - Fsin theta)
miu ( Fg - Fsin theta) = Fcos theta

i got up to here and idk how to solve it

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PhanthomJay
Homework Helper
Gold Member

## Homework Statement

You drag a suitcase of mass 17 kg with a
force of F at an angle 40.6◦ with respect to
the horizontal along a surface with kinetic
coefficient of friction 0.59.
The acceleration of gravity is 9.8 m/s2 .
If the suitcase is moving with constant ve-
locity 2.68 m/s, what is F?

Ff = miuFn
All forces equal

## The Attempt at a Solution

so first i broke it into components,
Ff = Fcostheta
Fn + Fsin theta = Fg
Ff = miuFn = miu( Fg - Fsin theta)
miu ( Fg - Fsin theta) = Fcos theta

i got up to here and idk how to solve it
You've done well. Just replace your 'Fg' with 'mg' so as not to confuse letter variables. Plug in the numbers and solve for F, it's all algebra from here.

cepheid
Staff Emeritus
Gold Member

Hi victoriee2, welcome to PF!

EDIT: PhantomJay beat me to it. I hope my elaboration is useful.

Guess what? You have basically solved this problem. All of the physics is done (meaning that you have applied all of the knowledge of the laws of motion that you needed in order to correctly set up the equations). Now it's just math. Algebra, to be more precise .

The nice thing is that you have only two unknowns: the applied force, F, and the normal force, FN. You also have two equations (for the sums of the forces in the horizontal and vertical directions, respectively). If the number of equations equals the number of unknowns, then a unique solution exists (i.e. the system is solvable). Practically speaking, this is because you can eliminate one of the unknowns by solving for it with one of the equations and substituting that result into the second equation. That is exactly what you have done. You used the first equation to eliminate FN, leaving only an equation in terms of F. Now all you have to do is solve that equation for F. You can do that by first distributing the 'mu' (which is how the name of that Greek letter is spelled btw) across the two terms on the left hand side. Then you can collect all of the terms involving F on one side of the equation and all of the other terms on the other side. After that, you just isolate F (i.e. do whatever division is necessary on both sides of equation to get it to look like "F = blah"). This is also called "solving" for F.

hey uhh thnx guys i got the answer for that part here is another part

If you are accelerating the suitcase with accel-
eration 1.39 m/s2 what is F?

For this part all i did was Fnet = ma
Since a is given i just plugged it in and since mass is constant i put Fnet = (17kg)(1.39m/s2) which is = to 23.63N, but it was wrong =[

PhanthomJay
Homework Helper
Gold Member

hey uhh thnx guys i got the answer for that part here is another part

If you are accelerating the suitcase with accel-
eration 1.39 m/s2 what is F?

For this part all i did was Fnet = ma
Since a is given i just plugged it in and since mass is constant i put Fnet = (17kg)(1.39m/s2) which is = to 23.63N, but it was wrong =[
F_net is 23.63 N, but the problem is asking for F, the force that you apply, not F_net. F_net = Fcostheta - Ff = ma = 23.63 N. You are going to have to solve for F using the approach similar to what you did before.

thnx alot man.

My next problem is
Two objects are connected by a string of neg-
ligible mass. The 12 kg block is placed on a
smooth table top 1.62 m above the floor, and
the 8 kg block hangs over the edge of the ta-
ble. The 8 kg block is then released from rest
at a distance of 0.81 m above the floor at time
t = 0.

i found the acceleration of the system already which is 3.92 m/s2

the question is,
How long dies it take the 8 kg block to strike
the floor?

i tried using the t = radical 2h/g but it gave me the wrong answer. i tried using the
dy = 1/2 (a)(t^2) where a = 9.8 + 3.92 but it gave me the wrong answer as well =[

PhanthomJay
Homework Helper
Gold Member

thnx alot man.

My next problem is
Two objects are connected by a string of neg-
ligible mass. The 12 kg block is placed on a
smooth table top 1.62 m above the floor, and
the 8 kg block hangs over the edge of the ta-
ble. The 8 kg block is then released from rest
at a distance of 0.81 m above the floor at time
t = 0.

i found the acceleration of the system already which is 3.92 m/s2

the question is,
How long dies it take the 8 kg block to strike
the floor?

i tried using the t = radical 2h/g but it gave me the wrong answer.
dy = 1/2 (a)(t^2) where a = 9.8 + 3.92 but it gave me the wrong answer as well =[
this one is right (assuming you plug in the proper number for dy), but since you know that a = 3.82m/s^2, why did you erroneously add the acceleration of gravity to it? The effect of gravity is already accounted for in the 8 kg block's weight; don't double count it.
Also, it is beneficial to post questions unrelated to the original topic in separate posts; you'll get a much better response that way, and it will avoid the confusion that sometimes develops when different problems are posted in a common post.

so i did dy = 1/2 g t^2

.81m = 1/2 (9.8)(t^2)
this comes out to be
t = .4065785562s

the answer was wrong though =[

PhanthomJay
Homework Helper
Gold Member

so i did dy = 1/2 g t^2

.81m = 1/2 (9.8)(t^2)
this comes out to be
t = .4065785562s

the answer was wrong though =[
You keep messing up your variables. dy=1/2(a)t^2. Why did you use 'g' instead of 'a'??? This is not a 'free fall' problem. The acceleration 'a' is what it is...a =3.92 m/s^2.

ohhh thnx