# M1 question- Resolving frictional forces

• Mar 15th 2010, 08:07 AM
fishkeeper
M1 question- Resolving frictional forces
Hi

I am having a problem with this question, and was wondering whether I could have a bit of a nudge in the correct direction?

I cannot do part 'iv'; however, I think that I may have worked out part iv for question iii though Im not sure

http://i213.photobucket.com/albums/c...00698Small.jpg

The answers I have so far are:

A) i)

Resolve horizontally- $\displaystyle 80*cos 25 = 72.5N$

ii)

Fmax= Coefficient of friction x R:

$\displaystyle Fmax= 0.32*80 = 25.6N$

iii)

Mass= $\displaystyle 80/9.8= 8.16kg$

-> $\displaystyle T + 8.16*cos 25 = 80*sin 25$

$\displaystyle T= 80*sin 25 - 8.16*cos 25 = 26.14N$

v)

$\displaystyle 80/9.8= 8.16kg$

Any help is greately appreciated concerning parts iii and iv
• Mar 15th 2010, 12:40 PM
skeeter
Quote:

Originally Posted by fishkeeper
Hi

I am having a problem with this question, and was wondering whether I could have a bit of a nudge in the correct direction?

I cannot do part 'iv'; however, I think that I may have worked out part iv for question iii though Im not sure

http://i213.photobucket.com/albums/c...00698Small.jpg

The answers I have so far are:

A) i)

Resolve horizontally- $\displaystyle 80*cos 25 = 72.5N$

ii)

Fmax= Coefficient of friction x R:

$\displaystyle Fmax= 0.32*80 = 25.6N$

iii)

Mass= $\displaystyle 80/9.8= 8.16kg$

-> $\displaystyle T + 8.16*cos 25 = 80*sin 25$

$\displaystyle T= 80*sin 25 - 8.16*cos 25 = 26.14N$

v)

$\displaystyle 80/9.8= 8.16kg$

Any help is greately appreciated concerning parts iii and iv

iii) $\displaystyle T_{min} = mg\sin{\theta} - f_{s \, max}$

iv) $\displaystyle T_{max} = mg\sin{\theta} + f_{s \, max}$
• Mar 15th 2010, 01:07 PM
fishkeeper
thankyou Skeeter

That helps enourmously!