A block of mass 20 kg is sitting on a surface inclined at = 45°. Determine the minimum force necessary to prevent slipping if the coefficient of static friction is 0.5.
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A block of mass 20 kg is sitting on a surface inclined at = 45°. If the coefficient of static friction is 0.5, then the minimum force necessary to prevent slipping is 35N .
Explanation:
Given: Mass (m) = 20 kg
Inclined angle = 45°
static friction = 0.5
the minimum force necessary to prevent slipping (F)
= mgsinθ − μmg cosθ
Take mg common
= mg (sinθ − μ cosθ)
Now put the values
F= (10)(9.8)(sin45°−0.5cos45°)
= 98 [1/ √2 - 0.5 / √2]
= 98[(1-0.5)/ √2]
= 98 [0.5/ √2]
F=35 N
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