Calculating Frictional Force & Normal Force: Coeff of Friction

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Homework Help Overview

The problem involves calculating the normal force and frictional force acting on a block subjected to an applied force at an angle. The context includes concepts of friction, normal force, and equilibrium in a physics setting.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculation of the normal force and frictional force, with one participant attempting to derive the normal force using the equation Fn = mg + 10sin30. There is a question regarding the calculation of the frictional force, with some participants suggesting the need to consider horizontal forces and free body diagrams.

Discussion Status

The discussion is ongoing, with participants exploring different aspects of the problem. Some guidance has been offered regarding the need to analyze horizontal forces, but there is no explicit consensus on the correct approach to calculating the frictional force.

Contextual Notes

Participants are working under the assumption that the block remains stationary, which influences the equilibrium conditions being discussed. There is also a reference to an answer key that presents a different value for the frictional force, prompting further inquiry.

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Homework Statement


A force of 10N is applied downward at an angle of 30 deg with respect to the horizontal on a block. There is friction between the block and the floor and the block remains stationary. M=2kg
(a) What is the magnitude of the normal force acting on the block?

(b) The coefficient of friction between the floor and block is μ = 0.8. What is the magnitude of the frictional force acting on the block?

Homework Equations


Fn = mg + 10sin30
Fn(mu) = Fk

The Attempt at a Solution


I got part A -- Fn = mg + 10sin30 = 19.6 + 5.0 = 24.6

But for part B, I thought it would be 24.6*.8 = 19.7, but according to the answer key, it should be 8.7.
 
Last edited:
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The block is stationary on the floor so the sum of the horizontal forces must equal 0. So draw the free body diagram along with the horizontal forces acting on it.
 
So, Fμ = 10 cos 30 ?
 
Correct.
 

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