# Forces-Newton's Laws and Vectors

• joseg707
In summary, the string makes an angle of 25 degrees with the vertical when the dice are accelerating.
joseg707

## Homework Statement

A pair of fuzzy dice is hanging by a string from your rearview mirror. While you are accelerating from a stoplight to 28 m/s in 6.0 s, what angle $$\vartheta$$ does the string make with the vertical?

## Homework Equations

$$\sum$$F=ma

## The Attempt at a Solution

I found the acceleration to be about 4.667, but I honestly have no clue where to go from here. I've looked for other examples in the book that are similar but I haven't found any. The angle part is really throwing me off. I'm completely lost and I just want to be pushed in the right direction.

Your value for the acceleration seems alright. Have you drawn a free-body diagram yet and carefully labeled all the forces acting on the dice? If you have, write it out here.

Try drawing a free body diagram

edit: jgens said it before me

Ok. Would it be something like this?

$$\uparrow$$ FT
[ ]$$\leftarrow$$ Acceleration?
$$\downarrow$$ FW

Not quite. Since, the dice are going to make an angle, Ft should not be positioned directly above mg; hence, your tension has two components, one in the x direction and one in the y. Additionally, your diagram should not show the acceleration as a force.

If it's easier you could just write out the results of your free-body diagram as:
∑Fx = ...
∑Fx = ...

FTcos$$\theta$$+FTsin$$\theta$$+FW.

Is that correct? Would i have to break it into two equations because of the x and y components?

Yes, it should be broken into two x and y components.

$$\sum$$Fx=FTsin$$\theta$$
$$\sum$$Fy=FTcos$$\theta$$+FW

Is that correct? I can substitute FW for mass times gravity right?

Your equations look correct and Fw can be replaced with mg (just make sure you use g < 0). However, you can extrapolate more still. Are the dice accelerating in the vertical component and what does this suggest about ∑Fy? Eventually, the dice must be accelerating at the same rate as the car, what does this suggest about ∑Fx (think about Newton's second law)?

Oh! So,

$$\sum$$Fy=0

because it is not accelerating, and

FTcos$$\theta$$+mg=ma

Would gravity be negative?

Well, it seems you've gotten your forces in the x and y directions mixed up. Your equations should read:

∑Fx = Ftsin(theta) = ma
∑Fy = Ftcos(theta) - abs(g)m = 0.

Oh yeah, sorry about that. But that still gives me three variables in both equations. I still only know that acceleration is 4.667m/s2 and gravity is 9.8m/s2.

True, but you can easily reduce it. Divide ∑Fy by ∑Fx and solve, or using ∑Fy come up with an expression for Ft and substitute it into ∑Fx and solve. Either method should work.

Ok, I got answer, 25 degrees, but I just want to make sure my math is correct.

I substituted Ftinto $$\sum$$Fx.

(mg/cos$$\theta$$)sin$$\theta$$=ma

mg*tan$$\theta$$=ma

tan$$\theta$$=a/g

tan-1(a/g)=$$\theta$$

Your math it correct. I can't check your angle measure because I don't have a calculator or table of values at this point in time but I would presume it is correct. Good job!

Thanks so much for your help! I've been having trouble understanding force and just going through that problem with you helped me a lot. Thanks so much for your time. :)

## 1. What are Newton's three laws of motion?

Newton's first law states that an object at rest will stay at rest and an object in motion will stay in motion with a constant velocity unless acted upon by an external force. Newton's second law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. Newton's third law states that for every action, there is an equal and opposite reaction.

## 2. How do forces affect an object's motion?

Forces can cause an object to accelerate, decelerate, or change direction. The magnitude and direction of the force will determine the resulting motion of the object.

## 3. What is the difference between scalar and vector quantities?

Scalar quantities have only magnitude, such as speed or temperature. Vector quantities have both magnitude and direction, such as velocity or force.

## 4. How do you calculate net force?

To calculate net force, you must add all of the individual forces acting on an object together. If the forces are acting in the same direction, you add them together. If they are acting in opposite directions, you subtract the smaller force from the larger one.

## 5. How do you resolve a vector into its components?

To resolve a vector into its components, you must use trigonometric functions to find the horizontal and vertical components of the vector. The horizontal component is the vector's magnitude multiplied by the cosine of its angle, and the vertical component is the vector's magnitude multiplied by the sine of its angle.

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