# Newtons Law and Tension

1. Feb 4, 2008

### jen333

1. The problem statement, all variables and given/known data
You are dragging a stack of shoeboxes along the floor by a rope (an odd situation I must say). If the total mass of the shoeboxes is 18kg and the rope is pulled at an angle of 45 degrees with no acceleration
i) what is N (normal force)? and
ii) What is the tension of there is a kinetic coefficient friction of 0.20?

2. The attempt at a solution

i) I know that N must be less than mg since the angle of the tension (rope) is carrying some of that weight.
$$\sum$$Fy=Wy+Fy+N=0
N=mg-Tsin45degrees
=(18kg)(9.81m/s$$^{2}$$)-Tsin45 degrees

Where I am stuck at is finding the tension in order to find N. or can i theoretically determine Tsin45 being half the mg since T=mg when the angle is 90 degrees.

ii) I have
mg(0,-1)+N(0,1)+T(cos45, sin45)+$$\mu$$kN(-1,0)

Ty=N-mg+Tsin45=0
Tx=Tcos45-$$\mu$$kN=0 (ma, when a=0)

to solve this, i would then isolate the Tsin45 and Tcos45 to solve for them, then use pythagoras' th. to find the total T.
but i'm not sure if this is the right way for solving it.

any help on especially part i is greatly appreciated! thanks

Last edited: Feb 4, 2008
2. Feb 4, 2008

### Shooting Star

> N=mg-Tsin45degrees =(18kg)(9.81m/s)+Tsin45 degrees

That is from the vertical forces.

> Where I am stuck at is finding the tension in order to find N.

From the horz forces, the other eqn would be

Tx = kN =k(mg - Ty), where Tx=Ty=T/(sqrt 2), k= 0.20.

You can find both N and T now.

3. Feb 4, 2008

### jen333

is there any way of determining N without k. Such as, if it were frictionless?

Last edited: Feb 4, 2008
4. Feb 4, 2008

### jen333

I've found an answer for ii

Ty=N-mg+Tsin45=0
N=mg-Tsin45

Tx=Tcos45-ukN=0
Tcos45=ukN

Substituting N into ukn:
Tcos45=uk(mg-Tsin45)
Tcos45=ukmg-ukTsin45

T= $$\frac{ukmg}{cos45-uksing45}$$
=$$\frac{(0.38)(18kg)(9.81)}{cos45-(0.38)(sin45)}$$
=153N

I hope this is the right method for Tension

but i'm still stuck for part i since it's asked first I'm assuming that the initial system is frictionless. (therefore, wouldn't the horizontal equation be zero leaving me stuck with only my initial vertical equation for N?)

5. Feb 4, 2008

### Shooting Star

The problem states, before i or ii that there is no accn. If there is a horz component Tx acting on the box, but there is no accn, there must be friction between the box and the floor to balance the Tx.

The box will undergo accn, be lifted off the ground, and N will vary until it vanishes.

I hope you understand now that it is not frictionless. The two eqns given in post #2 gives the complete solution. After that it's just algebra.