An object with a mass of $7 k g$ $7 kg$ is on a ramp at an incline of $\frac{π}{12}$ $pi/12$ . If the object is being pushed up the ramp with a force of $6 N$ $6 N$ , what is the minimum coefficient of static friction needed for the object to remain put?

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An object with a mass of $7 k g$ $7 kg$ is on a ramp at an incline of $\frac{π}{12}$ $pi/12$ . If the object is being pushed up the ramp with a force of $6 N$ $6 N$ , what is the minimum coefficient of static friction needed for the object to remain put?

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Answer 1 · 2016-02-02T14:22:46

$μ = (\frac{6}{m g cos (\frac{π}{12})} - tan (\frac{π}{12}))$ $mu = (6/(mgcos(pi/12)) - tan (pi/12))$

Explanation:

From FBD:
$F_{i} = F_{f} + F_{∣ ∣} = μ F_{N} + m g sin α$ $F_i = F_f + F_(|\|) = mu F_N + mgsinalpha$
$F_{i} = μ m g cos α + m g sin α$ $F_i = mumgcosalpha + mgsinalpha$
$6 = μ m g cos (\frac{π}{12}) + m g sin (\frac{π}{12})$ $6 = mumgcos(pi/12) + mgsin(pi/12)$
Solve for $μ$ $mu$
$μ = \frac{6 - m g sin (\frac{π}{12})}{m g cos (\frac{π}{12})}$ $mu = (6-mgsin(pi/12))/(mgcos(pi/12))$
$μ = (\frac{6}{m g cos (\frac{π}{12})} - tan (\frac{π}{12}))$ $mu = (6/(mgcos(pi/12)) - tan (pi/12))$
Now pull out your calculator and voila!

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An object with a mass of $7 k g$ $7 kg$ is on a ramp at an incline of $\frac{π}{12}$ $pi/12$ . If the object is being pushed up the ramp with a force of $6 N$ $6 N$ , what is the minimum coefficient of static friction needed for the object to remain put?

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Explanation:

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An object with a mass of 7kg7 kg is on a ramp at an incline of π12pi/12 . If the object is being pushed up the ramp with a force of 6N 6 N, what is the minimum coefficient of static friction needed for the object to remain put?

1 Answer

Explanation:

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An object with a mass of $7 k g$ $7 kg$ is on a ramp at an incline of $\frac{π}{12}$ $pi/12$ . If the object is being pushed up the ramp with a force of $6 N$ $6 N$ , what is the minimum coefficient of static friction needed for the object to remain put?