Co-efficient of Friction (Limiting Friction)

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The problem involves calculating the coefficient of friction for a 40kg box being pulled with a tension of 190N at a 33° angle. The normal force (R) is determined by resolving forces vertically, leading to the equation R = 392 - 190sin33°. The maximum frictional force (Fmax) is calculated using Fmax = 190cos33°. The coefficient of friction is then found by dividing Fmax by R, resulting in a value of approximately 0.55. Despite this calculation, the user reports that the online platform "MyMaths" indicates the answer is incorrect, prompting a request for assistance in identifying any errors in their method.
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Homework Statement


"A box of mass 40kg is pulled by a rope inclined at 33° to the horizontal. The box is about to slide and the tension in the rope is 190N. What is the co-efficient of friction (2 d.p.)?"

Homework Equations


Fmax = uR
g = 9.8 ms-2

The Attempt at a Solution


First, a diagram of the problem;

[PLAIN]http://img215.imageshack.us/img215/6131/m1stuck2.png

To find R, resolve upwards;

R + 190sin33° = 40*9.8
R = 392 - 190sin33°

Resolving horizontally to find Fmax;

Fmax = 190cos33°

So, the co-efficient of friction should be Fmax divided by R;

(190cos33°)/(392 - 190sin33°) = 0.552295128(...) = 0.55 (2 d.p.)

That is my total working for the problem, but the website I am using "MyMaths" is telling me it is incorrect, and I am assuming I must be incorrect because every time a new randomly-generated problem is created it tells me I am wrong when I solve that one using the method which I have described.

Could anyone please tell me the error in my method?

Thanks.
 
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I also get .55
 
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