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doesn't part C of this question contradict what was asked for in part b?

Question is attached. I accept part b has nothing wrong with it (solution attached) but part c clearly shows (part c solution is given as part of partb solution) that friction = 0.3 R = 0.3x(Mg+W) and so the mass does affect friction and so friction is not independent of the mass as stated in part b. I am totally split two ways by this question. Anyone who can help?

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Reply 1

Implication

Original post by thebrahmabull

I can't see the solution very clearly (maybe I'm getting old), but it looks like it says something like

F \leq 0.3R=0.3(Mg+W)

rather than

F = 0.3R

Reply 2

thebrahmabull

Original post by Implication

I can't see the solution very clearly (maybe I'm getting old), but it looks like it says something like

F \leq 0.3R=0.3(Mg+W)

rather than

F = 0.3R

Yes. But mew x R is friction by mathematical definition right?

(edited 8 years ago)

Reply 3

Implication

Original post by thebrahmabull

Yes. But mew x R is what friction is right?

\mu R

is the maximum force due to friction I believe. The actual force, denoted by

F

in this question, will be less than (or equal to) that.

I could be wrong though. It's been 4 years since I did A-level maths so I'm a bit fuzzy on this stuff :tongue:

Reply 4

username1560589

No. In B you were finding a formula for the friction force. In C you are applying a constraint on the friction.

Original post by thebrahmabull

Yes. But mew x R is friction by mathematical definition right?

No. Friction defined something like the force, parallel to the surface, that an object experiences due to interactions with the surface.

The case

F=\mu R

only applies when the object is moving or in limiting equilibrium.

Reply 5

thebrahmabull

Original post by Implication

\mu R

is the maximum force due to friction I believe. The actual force, denoted by

F

in this question, will be less than (or equal to) that.

I could be wrong though. It's been 4 years since I did A-level maths so I'm a bit fuzzy on this stuff :tongue:

Original post by morgan8002

No. In B you were finding a formula for the friction force. In C you are applying a constraint on the friction.

No. Friction defined something like the force, parallel to the surface, that an object experiences due to interactions with the surface.

The case

F=\mu R

only applies when the object is moving or in limiting equilibrium.

So should I always visualise that mew x R is something different from friction, e.g.its a horizontal force, but not friction?

(edited 8 years ago)

Reply 6

username1560589

Original post by thebrahmabull

So should I always visualise that mew x R is something different from friction, e.g.its a horizontal force, but not friction?

No. Visualise it as the maximum possible value of friction.

Reply 7

Implication

Original post by thebrahmabull

So should I always visualise that mew x R is something different from friction, e.g.its a horizontal force, but not friction?

\mu R

is the friction, but only in the limiting case i.e. when the object is moving or on the point of moving. Think of it as the maximum friction so

F_{\mathrm{max}}=\mu R

. The more you try to move something, the stronger the frictional force resisting the motion gets - until it reaches its maximum value, when the object begins to move. If the object isn't moving and you aren't told it's in limiting equilibrium, you have to find the friction another way.

Thanks guys!

Original post by morgan8002

F=\mu R

only applies when the object is moving or in limiting equilibrium.

Interesting question, I have read the thread but part B confuses me. How can friction exerted by ladder be independent of the mass of the brick? Since the ladder is not in equilibrium, it is now moving, so maximum friction must be acting on it... Mu x R value gets greater as mass of the brick increases till it becomes (3w/8) right ?