Finding actual static friction given mass, external force, and angle.

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SUMMARY

The discussion focuses on calculating the actual static frictional force acting on a block with a mass of 25.0 kg when a horizontal external force of 25.0 N is applied. The relevant equation for static friction is established as static frictional force = (coefficient of static friction)(mass)(gravity). The coefficient of static friction (µ) is determined to be 0.102, but the primary goal is to find the static frictional force, not the coefficient itself. The key takeaway is that the actual static frictional force must be less than or equal to µN, where N is the normal force.

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Scarlet-
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1. What is the actual static frictional force that acts on the block if an external force of 25.0 N acts horizontally on the block? ________N The mass of the block is 25.0kg. Assume g=9.80m/s^2



Homework Equations


I am using: static frictional force=(coefficient of static friction)(mass)(gravity)


The Attempt at a Solution


25N=us(25.0kg)(9.80m/s^2)
us=0.102

Of course this is obviously wrong because I am not searching for the coefficient of static friction. Would you be able to guide me to the correct equation for this type of problem?

4. Other equations I am given
T=fs, T=Mg, N=Mg


Thanks,
Scarlet-
 
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Is the block moving when the 25 N force acts on it?
 
Welcome to PF!

Hi Scarlet-! Welcome to PF! :smile:

(have a mu: µ and try using the X2 tag just above the Reply box :wink:)
Scarlet- said:
1. What is the actual static frictional force that acts on the block if an external force of 25.0 N acts horizontally on the block? ________N The mass of the block is 25.0kg. Assume g=9.80m/s^2

The question asks for the force, not for µ. :wink:

And remember, the actual static frictional force is less than or equal to µN.

If the block isn't moving, then the horizontal components must add to zero,

so the friction force must be … ? :smile:
 

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