What is the maximum frictional force?

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Homework Help Overview

The discussion revolves around calculating the maximum frictional force acting on a block on a 35-degree incline. The original poster expresses confusion regarding the calculations presented in a textbook, which differ significantly from their own results.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to calculate the frictional force using a sine function based on the incline angle, while questioning the textbook's use of a cosine function and the multiplication by gravitational acceleration. Some participants question the accuracy of the textbook's approach to weight and mass.

Discussion Status

Participants are actively discussing the discrepancies between the original poster's calculations and the textbook's answer. There is a recognition of potential confusion regarding the definitions of weight and mass, and some participants are considering the implications of these errors in the textbook.

Contextual Notes

The original poster references a specific weight of 120 N for the block, which is central to the calculations being discussed. There is an ongoing examination of the assumptions made regarding the forces acting on the block on the incline.

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
 

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