Mechanics Resolving Forces

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Goldenknight
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This is from the answer sheet
Shouldn't the direction of friction be in the direction of the deceleration?
Because even though its decelerating the direction of motion is to the right?

The workings show that positive direction taken to the right. But they show the acceleration as negative. And the frictional force also as negative so does not make sense here.

Was wondering if I am missing something?
Here is my diagram for this
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mqb2766
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Friction opposes motion (opposite of velocity), it doesnt matter whether the object is accelerating or decelerating. In this scenario it is the sole cause the deceleration so the frictional force is in the opposite direction to positive velocity..

For their diagram, Id guess the positive direction etc is to the left as friction is acting to the right. Similarly, a represents an arbitrary (positive) acceleration. For this example it turns out to be negative as it must equal the deceleration caused by the friction. When they use v-u+at, ithe sign of a is correctly worked out and it is decelerating as the question states. Note for a force diagram, once the positive direction is clearly stated, you'd mark on the force directions (and their magnitudes) relative to that.

Edit - as a related example, imagine you threw a ball up in the air and its motion was solely determined by gravity. Assuming positve is up, you'd mark an arbitrary positive acceleration a as upwards, g (9.8) acting downwards and youd get an equivalent diagram to what they have here. Then newton 2 gives
a = -9.8
so it is decelerating (upwards) or equivalently accelerating downwards.
Last edited by mqb2766; 1 month ago
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Goldenknight
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(Original post by mqb2766)
Friction opposes motion (opposite of velocity), it doesnt matter whether the object is accelerating or decelerating. In this scenario it is the sole cause the deceleration so the frictional force is in the opposite direction to positive velocity..

For their diagram, Id guess the positive direction etc is to the left as friction is acting to the right. Similarly, a represents an arbitrary (positive) acceleration. For this example it turns out to be negative as it must equal the deceleration caused by the friction. When they use v-u+at, ithe sign of a is correctly worked out and it is decelerating as the question states. Note for a force diagram, once the positive direction is clearly stated, you'd mark on the force directions (and their magnitudes) relative to that.

Edit - as a related example, imagine you threw a ball up in the air and its motion was solely determined by gravity. Assuming positve is up, you'd mark an arbitrary positive acceleration a as upwards, g (9.8) acting downwards and youd get an equivalent diagram to what they have here. Then newton 2 gives
a = -9.8
so it is decelerating (upwards) or equivalently accelerating downwards.
Ahh yes I got it now thank you! I actually put v and a in different directions which caused the problem. If I am taking right as positive then even a and v should be positive to the right
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mqb2766
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(Original post by Goldenknight)
Ahh yes I got it now thank you! I actually put v and a in different directions which caused the problem. If I am taking right as positive then even a and v should be positive to the right
The direction that you assume is positive on your sketch is obviously arbitrary. If you assumed positive (displacement) is right, then a positive velocity increases displacement and that is interpreted as pointing right. Similarly for positive acceleration. Drawing the frictional force muR would then be to the left as it acts to reduce a positive velocity and hence produces a negative acceleration (deceleration).

If the velocity was negative, friction would produce a positive acceleration and the velocity and friction arrows would be obviously reversed. They must be in opposing directions.

In your sketch in the OP, you have this, the only thing that might cause confusion is your "a". Im sure you mean that its the direction of the deceleration, so the acceleration is negative in the opposite direction to the current velocity. In the book, Im assuming they're referring to the direction of an arbitrary positive acceleration. For more complex problems, you may not be sure of whether the resultant force will accelerate or decelerate, so the main thing is to make sure the positive direction is clearly marked so the sign of individual forces can be determined correctly.
Last edited by mqb2766; 1 month ago
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