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M1 question please help :) thanks

Hi, I'm stuck on question iii because I don't get what they mean by least possible tension.
What equation do I have to form, please help thanks! :smile:

Reply 1

sac99

Ohh im awful at m1 :frown:

i need help with it, exam is quite soon as well. are you taking unit 2 physics?

Reply 2

Sokka

Original post by sac99

Ohh im awful at m1 :frown:

i need help with it, exam is quite soon as well. are you taking unit 2 physics?

na im in year 11 xD I agree M1 is kinda hard but trust me you need to understand the topics well to proceed. Like resolving forces for example, if you don't understand that, there's no way on earth you will do well in the slope/plane questions or pulleys and pegs questions. Everything in M1 has this little introductory topic you need to understand well. Also, if you don't understand vectors, you will have a hard time with projectiles. Anyways you doing AQA??

Reply 3

NovoStar93

I'm sorry that i cant be bothered to do the actual maths, however, least possible tension means the force that will keep the box in equilibrium with the 'down the slope' component of the weight of the box, remember to take account of friction, which in this case you can assume is a force pushing the box up the slope.

GOOD LUCK

Messed up the edit - Draw a force diagram it'll make it a lot easier.

(edited 12 years ago)

Reply 4

Sokka

Original post by NovoStar93

tyvm, but I don't get why the friction helps to push the box up. Please explain this, thanks mate :tongue:

Reply 5

dream123

I got 57N, is that correct?

Reply 6

The Bagel Guy

Original post by Sokka

tyvm, but I don't get why the friction helps to push the box up. Please explain this, thanks mate :tongue:

The friction always acts opposed to the direction of motion no? So it's not so much trying to push the box up so to speak, it's jsut stopping the box from moving. It just so happens to that to stop the box moving the frictional force is acting up the slope.

Not sure how clear that is.

Anyways, I'm a undergad mechanical engineer so If you have any other mechanics questions you can drop me a PM, hot high 90's in M1 and M2 and 100 in M3 so I know my stuff :wink:

Reply 7

NovoStar93

Hi, sorry if it's a bit late

Basically, friction doesn't push the box up really, it's just that it acts against the direction the weight of the box wants to pull it.

Eg/ if friction max is 50N but the weight of the box pulling down the slope is 60N then the actual force acting on the box down the slope is 10N.

Hope that helps.

Reply 8

ShahzaibMuneeb

Original post by Sokka

Hi, I'm stuck on question iii because I don't get what they mean by least possible tension.
What equation do I have to form, please help thanks! :smile:

Least possible tension is the least tension required for the block to stay in equilibrium, you can calculate the friction via 0.32 x 72.5 = 23.2 N (Friction < (or equal to) u x R) where u is "mu" meaning the coefficient of friction. So the tension need to match that friction PLUS the weight component down the slope which is 80sin(25) = 33.8. So we have to have a tension of friction + weight component down slope which is 23.2 + 33.8 = 57N, therefore 57N of tension in the rope is required to keep the block from moving down the slope (in a state of equilibrium).

(Friction is acting down the slope as the direction of motion is up the slope - well the intended direction of motion, in this model it's not moving)

Please do tell me what the markscheme says :smile:

(edited 12 years ago)