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A block of weight 15N slides on a horizo...

A block of weight `15N` slides on a horizontal table the co-efficient of siliding fricition is `0.4`. The area of the block in contact with the table is `0.5m^(2)`. The shearing stress will be

A

`120 Nm^(-2)`

B

`140 Nm^(-2)`

C

`160 Nm^(-2)`

D

`180 Nm^(-2)`

Text Solution

Verified by Experts

The correct Answer is:
A
Stress `= (mu mg)/(A) = F = f = mu mg`
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