Can't get correct sign for a simple force problem

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In summary, the force of kinetic friction is 1 N due to the block's acceleration of 0.750 m/s2 and the tension force of 2.50 N acting horizontally on the block. This can be calculated using the formula Fnet = ma = F1 + F2 + F3 + . . . and setting up the x axis in the direction of tension as positive and the opposite direction as negative. The force of tension is positive and the force of friction is negative, resulting in a positive value for the force of friction due to the use of absolute value and cos(180) instead of keeping the sign of the vector.
  • #1
Mr Davis 97
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Homework Statement


A tension force of 2.50 N acts horizontally on a 2.00 kg block. The block accelerates at 0.750 m/s2. What is the force of kinetic friction?

Homework Equations


Fnet = ma = F1 + F2 + F3 + . . .

The Attempt at a Solution


I define the reference frame to be, in terms of an x axis, positive to the right and negative to the left. Thus, the force of tension is positive because I define it to be in the right direction, and the force of friction should be negative since it opposes the tension.

##F_{fx}## = force of friction
##F_{tx}## = tension

##F_{netx} = ma = F_{tx} + F_{fx}##
##F_{fx} = ma - F_{tx}##
##\left \| \vec{F_{f}} \right \|cos(180^{\circ}) = ma - \left \| \vec{F_{t}} \right \|cos(0^{\circ})##
##-\left \| \vec{F_{f}} \right \| = ma - \left \| \vec{F_{t}} \right \|cos(0^{\circ})##
##\left \| \vec{F_{f}} \right \| = \left \| \vec{F_{t}} \right \|cos(0^{\circ}) - ma##
##\left \| \vec{F_{f}} \right \| = (2.50 ~N) - (2.00 ~kg)(0.750 ~m/s^{2})##
##\left \| \vec{F_{f}} \right \| = (2.50 ~N) - (1.50~N)##
##\left \| \vec{F_{f}} \right \| = 1~N##

As we can see, I get positive 1 N rather than -1 N. Why is this? What am I doing wrong?
 
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  • #2
Mr Davis 97 said:
I define the reference frame to be, in terms of an x axis, positive to the right and negative to the left. Thus, the force of tension is positive because I define it to be in the right direction, and the force of friction should be negative since it opposes the tension.
You don't need to say what is "left" and what is "right." For example, I would of set up the directions as: "the positive direction is the direction of tension and the negative direction is the direction opposite of tension." At any rate, this doesn't affect your answer.

Mr Davis 97 said:
##F_{fx} = ma - F_{tx}##
##\left \| \vec{F_{f}} \right \|cos(180^{\circ}) = ma - \left \| \vec{F_{t}} \right \|cos(0^{\circ})##
I'm not sure why you're doing this... But this step is why it turns out positive instead of negative. All you did in this step was multiply Ff by -1 (without multiplying the other side by -1) which is why your answer came out altered by a factor of -1

Mr Davis 97 said:
##\left \| \vec{F_{f}} \right \| = 1~N##

As we can see, I get positive 1 N rather than -1 N. Why is this? What am I doing wrong?
Anyway, doesn't the ||...|| mean "the magnitude of ... "? The magnitude of -1N is 1N.
 
  • #3
your absolute value Force is positive ... what direction does it push? (The sign of a vector tells its direction)
... I suggest you NOT use absolute values, and not use cos(180), so you get the F_x component with sign still attached.
 

FAQ: Can't get correct sign for a simple force problem

1. Why is it important to get the correct sign for a simple force problem?

The correct sign in a force problem indicates the direction of the force, which is crucial in determining the overall net force and the resulting motion of an object. Without the correct sign, the calculated force may be incorrect and lead to incorrect conclusions.

2. What are some common mistakes that lead to incorrect signs in force problems?

One common mistake is using the wrong coordinate system, which can result in assigning the incorrect positive and negative directions to forces. Another mistake is not considering the direction of the force relative to the object being analyzed, which can also lead to incorrect signs.

3. How can I determine the correct sign for a force in a simple force problem?

In order to determine the correct sign, you must first establish a coordinate system and assign positive and negative directions. Then, consider the direction of the force relative to the object and assign the appropriate sign based on whether the force is acting in the positive or negative direction.

4. What are some strategies for avoiding errors in determining the sign of a force?

To avoid errors, it is important to carefully establish a coordinate system and consistently use the same directions for positive and negative. Double-checking the direction of the force relative to the object and using diagrams or visual aids can also help prevent errors.

5. Are there any cases where the sign of a force does not matter?

Yes, in some cases, the sign of a force may not matter, such as when calculating the magnitude of the force or when analyzing the motion of an object in only one direction. However, it is still important to establish a coordinate system and assign directions to ensure accuracy and consistency in problem-solving.

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