What is the maximum frictional force?

AI Thread Summary
The discussion revolves around calculating the maximum frictional force of a block on a 35-degree incline. The user calculated it as 69N using sin(35) multiplied by 120N, while the book states it as 675N using a different formula. The confusion arises from the book's incorrect treatment of weight and mass, mistakenly multiplying the weight in Newtons by 9.81 instead of recognizing it as the force due to gravity. The user contemplates creating a Physics Forum page to correct such errors in textbooks. The conversation highlights the importance of accurately understanding the relationship between mass, weight, and gravitational force in physics calculations.
Shaye
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Homework Statement
The block in the diagram weighs 120N. It will start to slip down the slope when the component of it's weight acting down the slop equals the maximum frictional force. If the block slips when the angle Θ = 35 degrees what is the maximum frictional force?
Relevant Equations
Sin(Θ)*Opposite = Hypotenuse and F=Ma
In the attachment I am supposed to calculate the maximum frictional force of a block on a 35 degree angle incline (that is the point at which the force acting opposite frictional force is highest)

I make it out to be sin(35)*120N = 69N but the book says 675N and it gives it as cost(55)*120N*9.81. But given F=Ma how so?

I don't understand this or something has slipped my mind. The diagonal force component (hypotenuse) should be 69N because the veridical component is 120N.

Thanks
 

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The book has confused weight with mass in producing their answer. They've erroneously multiplied the weight in Newtons by 9.81, or g. One multiplies a mass by g to get force due to gravity. But they've provided that force already as a weight of 120 N.
 
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What book is this? I am thinking of creating a Physics Forum page of book corrections.
 
gneill said:
The book has confused weight with mass in producing their answer. They've erroneously multiplied the weight in Newtons by 9.81, or g. One multiplies a mass by g to get force due to gravity. But they've provided that force already as a weight of 120 N.

Phew! Thanks. I thought I was going mad
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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