Soul Wavel3ngth
Badges: 12
Rep:
?
#1
Report Thread starter 1 month ago
#1
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?
0
reply
casioFX991EX
Badges: 10
Rep:
?
#2
Report 1 month ago
#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
0
reply
Soul Wavel3ngth
Badges: 12
Rep:
?
#3
Report Thread starter 1 month ago
#3
(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
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?
0
reply
casioFX991EX
Badges: 10
Rep:
?
#4
Report 1 month ago
#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?
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).
0
reply
Soul Wavel3ngth
Badges: 12
Rep:
?
#5
Report Thread starter 1 month ago
#5
(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).
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
0
reply
casioFX991EX
Badges: 10
Rep:
?
#6
Report 1 month ago
#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
Yes you do exactly that which is what I showed in my original response - if this is what you are doing too surely you've calculated the same value?

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
0
reply
old_engineer
Badges: 11
Rep:
?
#7
Report 1 month ago
#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?
The minimum tension in the rope corresponds to the case where the box is trying to slide down the slope, and the rope (assisted by friction) is just about holding in stationary. Since the box is trying to slide down the slope, the frictional force must be opposing the attempted motion, and must therefore be acting UP the slope in this case. Thus Tmin + Fmax = 80sin(25).
1
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

Have you experienced financial difficulties as a student due to Covid-19?

Yes, I have really struggled financially (83)
18.78%
I have experienced some financial difficulties (125)
28.28%
I haven't experienced any financial difficulties and things have stayed the same (165)
37.33%
I have had better financial opportunities as a result of the pandemic (55)
12.44%
I've had another experience (let us know in the thread!) (14)
3.17%

Watched Threads

View All
Latest
My Feed