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Old Jun 8th 2009, 08:59 PM   #1
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Friction

I'm having difficulty with these questions can you please explain.

1) Explain why is fs less than or equal to us (coefficient of static friction)N?

2)Prove than tan ϑ is equal to uk (coefficient of kinetic friction) when a block slides down an incline with a constant speed. (use symbols, not numbers)
If the ϑ is the maximum angle of incline just before the block moves, what is us (coefficient of static friction) in terms of ϑ?

3)Suppose that a block were made to move up the inclined plane with uniform speed by suspending masses on a string over the pulley. Derive an equation for the coefficient of kinetic friction for this case in terms of suspended masses, the mass of the block, and the angle of decline. (Neglect any friction and mass effects of the pulley.)
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Old Jun 8th 2009, 09:58 PM   #2
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1) Explain why is fs less than or equal to us (coefficient of static friction)N?

The maximum value of the force of friction has been empirically determined to be equal to the product of the coefficient of static friction and the normal reaction. The force on a stationary object which is on a surface (which has friction) is gradually increased till it starts moving.

The force of friction opposes motion and arises only when there is an attempt to move the object. Otherwise if the max value of riction is present all the time, then the object would move without any force being applied! It increases proportionally to the applied force such that it cancels it (till the max value is reached). Thus it is always less than or equal to the force of friction.

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Old Jun 8th 2009, 10:43 PM   #3
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2) Prove than tan ϑ is equal to uk (coefficient of kinetic friction) when a block slides down an incline with a constant speed. (use symbols, not numbers)
If the
ϑ is the maximum angle of incline just before the block moves, what is us (coefficient of static friction) in terms of ϑ?

The block slides with a constant speed down the incline because the net force on it is zero. Thus, the component of its weight acting downward along the plane i.e. mg sinθ is balanced by the force of friction fs = μ k mg cosθ.
where mg cosθ is the normal reaction
Thus,
mg sinθ = μ k mg cosθ, or μ k = tanθ.

If a block is placed on an incline, and the angle of inclination is gradually increased, at a particular angle, the block just starts moving. Thus the force of static friction opposes the component of weight down the incline till this inclination is reached and this is the max value of the force of static friction.
Here also the above equations apply ; there is no motion because there is no net force, but the coefficient of friction used id the coefficient of static friction.

mg sinθ = μ s mg cosθ, or μ s = tanθ.s

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Old Jun 8th 2009, 11:17 PM   #4
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3)Suppose that a block were made to move up the inclined plane with uniform speed by suspending masses on a string over the pulley. Derive an equation for the coefficient of kinetic friction for this case in terms of suspended masses, the mass of the block, and the angle of decline. (Neglect any friction and mass effects of the pulley.)

Let the suspended mass be M, the mass on the incline be m and the tension in the string be T. We have
T Mg = Ma and T mg Sinθ - μ k mg cosθ = ma as the equations of motion for the two masses. Hence,

(T- Mg) / M = (T mg Sinθ - μ k mg cosθ ) / m.

Since the masses are connected by an inextensible sting, their accelerations have to be the same.
Solve for μ k .

In future, please post your attempts at solving problems and where exactly your problem lies.
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