Help with maths question please

Hi guys, so just stuck on part iv-working out the frictional force bit. Here's my working out for part iv):
F = ma
T2 -T1 -F = 5 x 0.5
27 -22 - F =2.5
5 -F =2.5
F=2.5
But F is meant to 5. Would really appreciate any help on where I went wrong.

Reply 1

sienna2266

Just attached here the full mark scheme: just stuck on part iv-working out the frictional force bit. Here's my working out for part iv):
F = ma
T2 -T1 -F = 5 x 0.5
27 -22 - F =2.5
5 -F =2.5
F=2.5
But F is meant to 5. Would really appreciate any help on where I went wrong.

Reply 2

old_engineer

Original post by sienna2266

It looks as though you've carried over the values of T1 and T2 from earlier parts of the question. But this is wrong as the acceleration of the system is now different. As a result, you have to reapply Fnet = ma to the 2kg and 3kg masses to find the new values of T1 and T2. Then you should be OK.

Reply 3

sienna2266

Original post by old_engineer

Thank you so much - so whenever the acceleration changes, the tension changes? If the table was rough to begin with and then the acceleration changes, would that mean that all the forces apart from weight would change - so tension(an internal force) and friction and all the other external forces would change?

Reply 4

sienna2266

Original post by old_engineer

Reply 5

old_engineer

Original post by sienna2266

Yes that's right. Looking at either of the suspended weights, ma = Fnet and Fnet = mg - T (or T - mg), so if "a" changes then T must also change as it is the only other variable in the equation.

Reply 6

sienna2266

Original post by old_engineer

Yes that's right. Looking at either of the suspended weights, ma = Fnet and Fnet = mg - T (or T - mg), so if "a" changes then T must also change as it is the only other variable in the equation.

Thank you so much-this makes sense! However, why doesn't the tension change here? This is someone else's post but also confused me as well :s-smilie:

https://www.thestudentroom.co.uk/showthread.php?t=5202828

As discussed, when the acceleration changes then the tension and all other forces apart from weight changes, so why does it not apply to the question in the other person's thread?

(edited 6 years ago)

Reply 7

old_engineer

Original post by sienna2266

Thank you so much-this makes sense! However, why doesn't the tension change here? This is someone else's post but also confused me as well :s-smilie:

The short answer is that there is another force involved, namely resistance. Fnet = T - mg - R, and it is the introduction of R that causes the change in acceleration.

Reply 8

sienna2266

Original post by old_engineer

The short answer is that there is another force involved, namely resistance. Fnet = T - mg - R, and it is the introduction of R that causes the change in acceleration.

But in my question which is about the table and the pulleys, it is the introduction of friction which cause a change in acceleration (cause if the table is not smooth,then friction is what changes the acceleration). Similarly, on the other person's question, it is the introduction of resistance R which changes the acceleration. However, in my question, the tensions are changed but in the other question the tensions are not changed. Which brings me back to square one :frown:

(edited 6 years ago)

Reply 9

old_engineer

Original post by sienna2266

That's a good question. The answer, in this case, is that the frictional force is not acting on the suspended weights. The components of Fnet for the suspended weights remain as just T and mg. Therefore if a changes, so must T change.