4
$\begingroup$

A 1370-kg car is skidding to a stop along a horizontal surface. The car decelerates from 27.6 m/s to a rest position in 3.15 seconds. Assuming negligible air resistance, determine the coefficient of friction between the car tires and the road surface.

source: https://www.physicsclassroom.com/calcpad/newtlaws/prob30.cfm

I have got this question here but i am a bit stuck when i got force friction is equal to force applied which is $F=ma$. Acceleration appears to be negative and as a result, the force applied is a negative value. What does this mean? I know that negative involves direction but i still do not fully understand how to use the negative sign. So do we always assume the object is going to the right or in x-axis? If that so, f=negative doesn't make sense. Does that also mean that the friction is positive and going right?

Improve this question
$\endgroup$
1
  • $\begingroup$ Let us consider that the car is going rightwards. A negative force(or acceleration) would imply that the force is acting leftwards. The negative of a vector has the same length but opposite direction. $\endgroup$
    – Harshit Joshi
    Commented Sep 3, 2018 at 12:39

2 Answers 2

Reset to default
3
$\begingroup$

Positive and negative depends on your chosen coordinate system. When you choose a direction to be "positive", you're choosing a coordinate system. Acceleration is the time derivative of velocity and is a vector. In a one-dimensional problem, its direction is uniquely determined by its sign (there are only 2 possible directions; positive and negative). It may look like a scalar in this case, but it really is a vector. When acceleration is positive, it tells us that the velocity is moving to the right on the number line.

Negative acceleration, on the other hand, means that the object's velocity is decreasing (either slowing down in the positive direction or speeding up in the negative direction) i.e. moving to the left on the number line. It does not tell us how close to zero on the number line the velocity is. That is the job of the term "speed", which tells us the magnitude of the velocity, how far it is from zero. Both positive and negative acceleration can cause an object to speed up (in the respective directions). An object moving to the left and speeding up will have a very different motion compared to an object moving to the right and slowing down, yet both have would negative acceleration as their velocities are both moving to the left on the number line. Be careful when distinguishing "decreasing velocity" with "slowing down"; the former talks about velocity, while the latter talks about the speed.

In summary, acceleration itself tells us nothing about the velocity at which the object is already traveling at, it only tells us how fast and in which direction its velocity is changing. Hence the constant term when acceleration is integrated with respect to time.

Improve this answer
$\endgroup$
10
  • $\begingroup$ Your answer is okay until your last sentence. A negative sign on a force always means that the force is acting in the defined negative coordinate direction. If the object's velocity is also negative, the force does NOT opposed the motion. And the phrase "in general" is not applicable to negative signs on forces. $\endgroup$
    – Bill N
    Commented Sep 3, 2018 at 12:56
  • 1
    $\begingroup$ @Bill N: You're right, fixed. $\endgroup$
    – Vincent Thacker
    Commented Sep 3, 2018 at 12:57
  • $\begingroup$ Re. "the term "speed", which tells us the magnitude of the velocity, ". Then 'speed' is an unsigned positive number?? $\endgroup$
    – user45664
    Commented Sep 3, 2018 at 16:49
  • $\begingroup$ @user7777777 so what does negative force means? Does it means force is decereasing? and how do i post my response ? $\endgroup$
    – Jon Wick
    Commented Sep 4, 2018 at 2:34
  • $\begingroup$ @user7777777 so in a free body diagram, force weight is going in -y-axis direction does that mean force normal is positive? but you cant get a negative mass or negative gravity then force normal is negative? $\endgroup$
    – Jon Wick
    Commented Sep 4, 2018 at 2:38
1
$\begingroup$

I could write a force $\vec G$ as a vector $\vec G = G\,\hat g$ where $\hat g$ is the unit vector in the direction of the force and $G$ is the magnitude of the force.
In this equation $G$ is always positive because it is the magnitude of the force.

Now suppose that a force $\vec F$ acts parallel to the x-axis and the unit vector in the direction of $x$ increasing is $\hat x$.
In this case the force can be written as $\vec F = F \,\hat x$ where $F$ is called the component of the force in the $\hat x$ direction.
The important difference from the previous notation for the force $\vec G$ is that now the component $F$ can either be a positive or negative value.

A force $3\, \hat x$ has a component $3$ in the $\hat x$ direction ie it is a vector of length (magnitude) $3$ in the positive x-direction.
A force $-5\, \hat x=5(-\hat x)$ has a component $-5$ in the $\hat x$ direction ie it is a vector of length (magnitude) $5$ in the negative x-direction.

—-

Your Newton’s second law equation is a vector equation $\vec F = m\, \vec a$ which can be written as $F \hat x = m\, a \hat x$ where $F$ and $a$ are components of the force and the acceleration in the $\hat x$ direction (positive x-direction).
Removing the unit vectors which occur on both sides toilets $F=ma$.

Now suppose the velocity is $7\hat x$.
A deceleration (reduction in magnitude of the velocity) means that the acceleration is in the opposite direction to the velocity and perhaps might be $-2\hat x$.
Using $F=ma$, remember this comes from $F\hat x = m \hat x$, you get $F = m (-2) = -2 m$ which tells you that the force acts in the negative x-direction (-\hat x) ie opposite to the direction of the velocity which is in the positive x-direction.

If the initial velocity had been $-7\hat x$ (in the negative x-direction) then a deceleration might be $2\hat x$ (in the positive x-direction).

Improve this answer
$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.