A block of mass 100 kg is being moved steadily by pulling on a rope at

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SUMMARY

The discussion centers on calculating the coefficient of friction (μ) for a 100 kg block being pulled with a tension of 120 N at a 30-degree angle. The initial calculations yielded μ = 0.141, while the expected answer is μ = 0.113. Key equations used include T - Ff = ma and F = μR. The correct approach involves calculating the gravitational force, the vertical component of tension, and ensuring all forces are accurately accounted for in the friction calculation.

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Homework Statement

a block of mass 100 kg is being moved steadily by pulling on a rope attached to the block making an angle of 30 degree with the horizontal. if the tension of the rope is 120 N, what is the coefficient of friction?

Homework Equations


T- Ff = ma
F= μR

The Attempt at a Solution


R= mg cos x = 850 N
mg sin x = 490 N
a = 0, F= Ff
Ff = 120 N
F= μR
120= μ(850)
μ= 0.141
* ( answer given μ= 0.113. where is my mistake?)
 
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jigoku_snow said:

Homework Statement




a block of mass 100 kg is being moved steadily by pulling on a rope attached to the block making an angle of 30 degree with the horizontal. if the tension of the rope is 120 N, what is the coefficient of friction?

Homework Equations


T- Ff = ma
F= μR


The Attempt at a Solution


R= mg cos x = 850 N
mg sin x = 490 N
a = 0, F= Ff
Ff = 120 N
F= μR
120= μ(850)
μ= 0.141
* ( answer given μ= 0.113. where is my mistake?)

Not sure how/why you are doing any of those calculations. It always helps to say what you are about to do, then show the calculation.

Like for example:
Calculating force of gravity on the block : 100 x 9.8 = 980 N

Calculating vertical component of Tension: 120 x sin(30) = 60N

and what ever else you think is appropriate.

You then connect what you are calculating to give the answer you are after. A diagram can help.
 

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