2 crates shown in the figure below are on a horizontal frictionless surface. The woman is wearing golf shoes for traction as she applies a horizontal force of F= 50.0 N to the 6.00kg crate. The ropes are of negligible mass. Smaller box weighs 4.00kg
a) draw a free body diagram for each of the crates and the woman and for each force, indicate what body exerts that force.
b) What is the acceleration of the 6.00kg crate?
c) what is the tension T in the rope connecting the 2 crates.
drawing shows
o-T--0--->person
(the first is a small box and in between the bigger box is T for tension and the arrow is the force direction without a angle..basically directly horizontal to the x axis if I can describe it like that.
~my thinking..
a.) not sure about the woman's free body diagram
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b.)
F= 50N
a=?
m= smaller box + larger box = 4.00kg + 6.00kg= 10.00kg= 10,000g
F= ma
50N= 10,000g*a
a= 5e-3N/g
Is this fine?
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c.)
How do I find the tension in the rope between the 2 boxes?
~Thank you~
the woman has a force pulling her back (ie, the boxes are pulling back with 50N, and the shoes on the floor are pulling forward 50N.)
When doing F=ma, M is in kg
a=50/10= 5m/s^2
The acceleration of the smaller box is the same, 5m/s^2, so T= ma=4kg*5m/s^2=20N
Oh okay Thanks Bob
a) To draw the free body diagram for each of the crates and the woman, you need to consider all the forces acting on them.
For the smaller box:
- There is a force due to gravity acting downwards (mg, where m is the mass and g is the acceleration due to gravity).
- There is a tension force in the rope pulling upwards (T), which is connected to the larger box.
- There is also a normal force (N) acting upwards, equal in magnitude and opposite in direction to the force of gravity.
For the larger box:
- There is a force due to gravity acting downwards (mg).
- There is a force exerted by the woman in the horizontal direction (F).
- There is a tension force in the rope pulling towards the woman (T).
- There is a normal force (N) acting upwards, equal in magnitude and opposite in direction to the force of gravity.
For the woman:
- There is a force exerted by the woman on the larger box in the horizontal direction (F).
b) To find the acceleration of the 6.00kg crate, you can use Newton's Second Law (F = ma).
- The force acting on the 6.00kg crate is the horizontal force exerted by the woman (F = 50.0 N).
- The mass of the 6.00kg crate is given as 6.00kg.
Substituting these values into the equation F = ma and solving for acceleration (a), you get:
a = F/m = 50.0 N / 6.00kg = 8.33 m/s^2
c) To find the tension (T) in the rope connecting the two crates, you can use Newton's Second Law (F = ma).
- The force acting on the larger box is the horizontal force exerted by the woman (F = 50.0 N).
- The mass of the larger box is given as 6.00kg.
- The total mass of both crates is given as 10.00kg.
- The acceleration of the system (both boxes) can be found as the same value calculated in part (b) above (a = 8.33 m/s^2).
Considering the larger box separately:
F - T = m*a
Substituting the values, you get:
50.0 N - T = 6.00kg * 8.33 m/s^2
Solving for T, you can find the tension in the rope connecting the two crates.