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https://pmt.physicsandmathstutor.com...g%20Forces.pdf
The question I need help on is the very first one, part iii).
So my value for the maximum frictional force is roughly 23.2N however I'm not sure how I'd go about calcing the least possible tension in the rope, do I make the tension equal to the max frictional force + horizontal component of the box's weight (80sin25) and then solve from there?
The question I need help on is the very first one, part iii).
So my value for the maximum frictional force is roughly 23.2N however I'm not sure how I'd go about calcing the least possible tension in the rope, do I make the tension equal to the max frictional force + horizontal component of the box's weight (80sin25) and then solve from there?
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#2
T + The Max Frictional Force = The weight component parallel to the slope
Max Frictional Force = 23.2N
The weight component parallel to the slope = 80sin(25) = 33.8N
=> T = 33.8 - 23.2
Therefore, T = 10.6N
Max Frictional Force = 23.2N
The weight component parallel to the slope = 80sin(25) = 33.8N
=> T = 33.8 - 23.2
Therefore, T = 10.6N
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(Original post by casioFX991EX)
T + The Max Frictional Force = The weight component parallel to the slope
Max Frictional Force = 23.2N
The weight component parallel to the slope = 80sin(25) = 33.8N
=> T = 33.8 - 23.2
Therefore, T = 10.6N
T + The Max Frictional Force = The weight component parallel to the slope
Max Frictional Force = 23.2N
The weight component parallel to the slope = 80sin(25) = 33.8N
=> T = 33.8 - 23.2
Therefore, T = 10.6N
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#4
(Original post by Soul Wavel3ngth)
doesn't the weight component parallel to the slope usually add up with the max frictional force when it comes to slopes? Does it not apply here here due to it being in equilibrium?
doesn't the weight component parallel to the slope usually add up with the max frictional force when it comes to slopes? Does it not apply here here due to it being in equilibrium?
However, the resistive force due to friction would vary between both slopes and so it would be easier to pull the object up the slope with a lower coefficient of friction (i.e. the value of T would be lower for that slope).
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(Original post by casioFX991EX)
No they are seperate forces. Imagine two slopes, one with with a high coefficient of friction and one with a low coefficient of friction. If you placed the same object on both, the weight component parallel to the slope would be the same as the weight is the same.
However, the resistive force due to friction would vary between both slopes and so it would be easier to pull the object up the slope with a lower coefficient of friction (i.e. the value of T would be lower for that slope).
No they are seperate forces. Imagine two slopes, one with with a high coefficient of friction and one with a low coefficient of friction. If you placed the same object on both, the weight component parallel to the slope would be the same as the weight is the same.
However, the resistive force due to friction would vary between both slopes and so it would be easier to pull the object up the slope with a lower coefficient of friction (i.e. the value of T would be lower for that slope).
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#6
(Original post by Soul Wavel3ngth)
So even if the frictional force (diagonally downwards) and the weight component parallel to the slope (also diagonally downwards) act in the same direction you don't add them and make it equal to T? Hmmm it's weird since that's the way I've been taught it for a while now, I'll try your method though, thanks
So even if the frictional force (diagonally downwards) and the weight component parallel to the slope (also diagonally downwards) act in the same direction you don't add them and make it equal to T? Hmmm it's weird since that's the way I've been taught it for a while now, I'll try your method though, thanks
I'm sorry if I misinterpreted your second query but I thought you were implying that the resistive force is already included within the weight component which is all I was trying to address in my second response. Either way, I'm just trying to help
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#7
(Original post by Soul Wavel3ngth)
doesn't the weight component parallel to the slope usually add up with the max frictional force when it comes to slopes (since they're both acting in the same direction)? Does it not apply here here due to it being in equilibrium?
doesn't the weight component parallel to the slope usually add up with the max frictional force when it comes to slopes (since they're both acting in the same direction)? Does it not apply here here due to it being in equilibrium?
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