The total normal force (F) acting on the area of contact of disc clutch (Assuming uniform distribution of interface pressure) is:(Where P is the intensity of pressure, r0 and ri outer and inner diameter of disc respectively)

Question

TuteeHUB · Accepted Answer

$F = \pi P{\left( {r_0^4 - r_i^4} \right)^{1/2}}$

TuteeHUB · Answer

$F = \pi P\left( {r_0^2 - r_i^2} \right)$

TuteeHUB · Answer

$F = 2\pi P{r_i}\left( {{r_0} - {r_i}} \right)$

TuteeHUB · Answer

$F = 2\pi P{r_0}\left( {{r_0} - {r_i}} \right)$

1.	The total normal force (F) acting on the area of contact of disc clutch (Assuming uniform distribution of interface pressure) is:(Where P is the intensity of pressure, r0 and ri outer and inner diameter of disc respectively)
A.	\(F = \pi P\left( {r_0^2 - r_i^2} \right)\)
B.	\(F = \pi P{\left( {r_0^4 - r_i^4} \right)^{1/2}}\)
C.	\(F = 2\pi P{r_i}\left( {{r_0} - {r_i}} \right)\)
D.	\(F = 2\pi P{r_0}\left( {{r_0} - {r_i}} \right)\)
Answer» B. \(F = \pi P{\left( {r_0^4 - r_i^4} \right)^{1/2}}\)

The total normal force (F) acting on the area of contact of disc clutch (Assuming uniform distribution of interface pressure) is:(Where P is the intensity of pressure, r0 and ri outer and inner diameter of disc respectively)

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