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

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A block with a mass of 100 kg is being pulled with a rope at a 30-degree angle, and the tension in the rope is 120 N. The calculations for the normal force and frictional force were performed, leading to a coefficient of friction (μ) of 0.141, which differs from the expected value of 0.113. The discrepancy may stem from incorrect calculations of the vertical component of tension and the forces acting on the block. It is suggested to clearly outline each calculation step and consider using a diagram for clarity. Accurate calculations and a systematic approach are essential for determining the correct coefficient of friction.
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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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