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4 years ago

15

A 60-N box rests on a rough horizontal surface with a coefficient of static friction of 0.5. A horizontal force of 23 N acts on

the box but the box is observed to be at rest. What is the value of the static friction force

1 answer:

Ira Lisetskai [31]4 years ago

4 0

Answer:23 N

Explanation:

Given

Weight of box $W=60\ N$

Coefficient of static friction is $\mu _s=0.5$

Applied force $F=23\ N$

When Force is applied box is observed to be at rest i.e. static friction is overcoming the applied force.

Thus Static friction $F_s$ =applied force

$F_s=23\ N$

Although it maximum value can go up to $(F_s)_{max}=\mu _sN$

$F_s=0.5\times 60$

$F_s=30\ N$

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4 years ago

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jeyben [28]

Here i hope this helps, let me know if you need clarification on anything :)

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