Solving the Forces on a Box in a Moving Car

In summary, a box with mass m, held by two horizontal ropes and two vertical ropes, is inside a still car. If the car accelerates suddenly with the acceleration of a, the box will remain still inside the car. The acceleration of the box/car can be calculated using the equations ma + t1 = t3 and mg + t2 = t4, which results in a = g(t3 - t1)/(t4 - t2). The separation after time t can be calculated using the equation s = \frac{gt^2(t3-t1)}{2(t4-t2)}, assuming the starting velocity of the car is 0. Tension forces always pull away from the object they act on.
  • #1
Hi, I'm new here. I have a problem.

Homework Statement


A box has the mass m, held by two horizontal ropes and two vertical ropes. The box is inside a still car. If the car accelerated suddenly with the accekeration of a then the box will remain still inside the car. What is :
a. The acceleration of the box/car? (state it in t1, t2, t3, t4 and g)
b. The separation after the time t? (state it in t1, t2, t3, t4, t and g)

The illustration is here -> xttp://i31.tinypic.com/izvnev.jpg (change x to h)

Homework Equations


[tex]\Sigma F = ma[/tex]

The Attempt at a Solution



a.
t3 + t1 = ma
m = (t3 + t1)/a (1)

t4 + t2 = mg
m = (t4 + t2)/g (2)

(t3 + t1)/a = (t4 + t2)/g
a = g(t3 + t1)/ t4 + t2 (3)

b.
S = Vt
S = a * t * t
S = (g(t3 + t1)/ t4 + t2) * t^2
S = g (t^2) ((t3 + t1)/(t4 + t2))I actually don't really know how the tensions work. I hope someone can explain it to me. And sorry, I'm not used to using latex yet.
 
Last edited:
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  • #2
Vermillion said:
Hi, I'm new here. I have a problem.

Homework Statement


A box has the mass m, held by two horizontal ropes and two vertical ropes. The box is inside a still car. If the car accelerated suddenly with the accekeration of a then the box will remain still inside the car. What is :
a. The acceleration of the box/car? (state it in t1, t2, t3, t4 and g)
b. The separation after the time t? (state it in t1, t2, t3, t4, t and g)

The illustration is here -> xttp://i31.tinypic.com/izvnev.jpg (change x to h)

Homework Equations


[tex]\Sigma F = ma[/tex]


The Attempt at a Solution



a.
t3 + t1 = ma
m = (t3 + t1)/a (1)

t4 + t2 = mg
m = (t4 + t2)/g (2)

(t3 + t1)/a = (t4 + t2)/g
a = g(t3 + t1)/ t4 + t2 (3)

b.
S = Vt
S = a * t * t
S = (g(t3 + t1)/ t4 + t2) * t^2
S = g (t^2) ((t3 + t1)/(t4 + t2))


I actually don't really know how the tensions work. I hope someone can explain it to me. And sorry, I'm not used to using latex yet.
Hi, Vermillion, and welcome to the Forums! A few points here to help you along in part (a), if I understand the problem correctly, and assume horizontal motion and massless, inextensible ropes:: First, Tension forces always pull away from the object they act on, so in your equations, check your plus and minus signs. Secondly, you have equated your 2 equations when solving each for m, which is unnecessary. The acceleration in the horizontal and vertical directions are independent of each other. Thirdly, with the box accelerating to the right (with respect to the ground), the box would appear to fare quite well without the left and bottom ropes present.

For part (b), what separation is the problem talking about? Since the box remains still inside the car, there can be no separation between the box and car, so I guess the problem means to ask about the displacement of the box with respect to the ground, in which case you need to use your basic motion eqations.
 
  • #3
Oh wow thanks! I actually never knew that tension forces always pull away from the object they act on, duh... As for the 2 equations of m, i needed them to solve the a(acceleration) in t1, t2, t3 ,t4, and g. My third equation is actually m = m. Sorry, i didn't type it.

I think i have done the correct equations now.
ma + t1 = t3
ma = t3 - t1
m = (t3 - t1)/a (1)

mg + t2 = t4
mg = t4 - t2
m = (t4 - t2)/g (2)

m = m
(t3 - t1)/a = (t4 - t2)/g
a = g(t3 - t1)/(t4- t2 (3)

So the separation would be

s = vt
s = a*t^2
s = (t^2*g(t3-t1)) / t4-t2

Is that correct?
 
  • #4
Vermillion said:
Oh wow thanks! I actually never knew that tension forces always pull away from the object they act on, duh... As for the 2 equations of m, i needed them to solve the a(acceleration) in t1, t2, t3 ,t4, and g. My third equation is actually m = m. Sorry, i didn't type it.

I think i have done the correct equations now.
ma + t1 = t3
ma = t3 - t1
m = (t3 - t1)/a (1)

mg + t2 = t4
mg = t4 - t2
m = (t4 - t2)/g (2)

m = m
(t3 - t1)/a = (t4 - t2)/g
a = g(t3 - t1)/(t4- t2 (3)

So the separation would be

s = vt
s = a*t^2
s = (t^2*g(t3-t1)) / t4-t2

Is that correct?
Part (a) is correct, since the problem asked you to solve for the acceleration in terms of those variables. Part (b), however, is not correct. In the equation s=vt, the 'v' refers to the average velocity (the average of the initial velocity and final velocity at time 't'), whereas in the equation V=at, the 'V' refers to the final velocity at the end of the time period of an object starting from rest. Average velocity and final velocity are not the same. How are they related?
 
  • #5
You're right, i forgot the velocity is not constant. Then i should use

[tex]s = v0t + \frac{at^2}{2}[/tex]

Because the car was resting then

[tex]v0 = 0[/tex]

[tex]s = \frac{at^2}{2}[/tex]

[tex]s = \frac{gt^2(t3-t1)}{2(t4-t2)}[/tex]

Is that correct?
 
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  • #6
Yes, where 's' represents the distance the box (and car) have moved together from their original starting position with respect to the ground. I'm not sure what the problem means by 'separation', we're assuming it means 'displacement'.
 
  • #7
Yeah i think that's what the question meant. Thanks for the help! :)
 

1. What forces act on a box in a moving car?

There are two main forces acting on a box in a moving car: the force of gravity and the force of inertia. The force of gravity pulls the box downward, while the force of inertia keeps the box moving in a straight line at a constant speed, unless acted upon by an external force.

2. How does the force of gravity affect the box?

The force of gravity acts downward on the box and causes it to have a weight. This weight is equal to the mass of the box multiplied by the acceleration due to gravity (9.8 m/s² on Earth). The force of gravity also determines the normal force exerted by the car's surface on the box.

3. What is the force of inertia and how does it affect the box?

The force of inertia is the tendency of an object to resist changes in its motion. In this case, the box has a force of inertia that keeps it moving at a constant speed in a straight line, even as the car turns or accelerates. This force of inertia is proportional to the mass of the box and the acceleration of the car.

4. How does the direction of the car's movement affect the forces on the box?

The direction of the car's movement affects the forces on the box in two ways. First, if the car is accelerating or decelerating, the force of inertia will change and affect the box accordingly. Second, if the car is turning, the direction of the forces acting on the box will change, as the normal force exerted by the car's surface will shift.

5. How can the forces on a box in a moving car be calculated?

To calculate the forces on a box in a moving car, you can use Newton's second law of motion, which states that force is equal to mass multiplied by acceleration (F=ma). By considering the mass of the box and the acceleration of the car, you can determine the force of inertia acting on the box. Additionally, by considering the force of gravity and the normal force exerted by the car's surface, you can calculate the net force on the box.

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