Using newton laws

1. Jul 19, 2007

Edwardo_Elric

1. The problem statement, all variables and given/known data
The crates are on a horizontal frictionless surface. The woman (wearing golf shoes so that she can get traction) applies a horizontal force F = 50.0N to the 6.00kg crate.
a.) what is the magnitude of the acceleration of the 6.00kg crate?
b.) what is the tension T in the rope (of negligible mass) connecting the two crates?

2. Relevant equations
$$F = ma$$

3. The attempt at a solution

a.) since the horizontal component of force has only 50 newtons with zero friction:
$$F_x = ma$$
$$a = \frac{m}{F_{x}}$$
$$a = \frac{6.00kg}{50.0N}$$
$$a = 8.33 \frac{m}{s^2}$$

b.) Tension?
No other x component of force
so i believe the tension is 50.0N

Last edited: Jul 19, 2007
2. Jul 19, 2007

HallsofIvy

Staff Emeritus
But she's not moving only the 6 kg crate! Yes, the 50 N force is on the 6 kg crate but there is also a force (that you don't know yet) caused by the drag of the 4 kg chart.

In essence she is moving both crates: a total of 10 kg. There common acceleration is
[tex]\frac{50}{10}= 5 m/s^2[/itex]
(you have your fraction upside down! Since F= ma, a= F/m, not m/F.)

Now, what what NET force would produce that acceleration in the 6 kg crate alone? (F= ma= (6)(5)) The difference between that and the 50 N is the "drag" of the 4 kg crate and is the tension in the rope between.

3. Jul 19, 2007

ank_gl

you have the acceleration of both the blocks. draw the free body diagram of the two blocks saperately and find the tension

4. Jul 19, 2007

ank_gl

hey pretty much beautiful drawing of the lady there:)

5. Jul 19, 2007

Edwardo_Elric

hey thanks a lot
so 20N is the tension between the crate

i think when you already have an acceleration you can instantly compute for the tension between 4kg and 6kg mass... using the 4kg mass,

so 4kg(5m/s^2) = 20N

Last edited: Jul 19, 2007
6. Jul 20, 2007

ank_gl

right on. the tension is the only force moving the 4kg mass.

7. Jul 22, 2007

Edwardo_Elric

w8 how would i draw a free body diagram on the two blocks including the woman? and the forces acting on each body?

8. Jul 22, 2007

ank_gl

F is 50N from the problem, you have the acceleration of the blocks(that is 5m/s/s).
only force on 4kg block is tension in the rope.
on the 6kg block, net force in the direction of acceleration or motion is applied force - tension in the rope.
on the lady, friction balances the 50 N