A block of wood weighs 10 N and is resting on an inclined plane. The coefficient of friction is 0.7. What will be the frictional force that acts on the block when the plane is 30 degrees inclined with the horizontal?
vasudevrathod:
sorry i dont know anything about physics it is going to much tough to me.
but i will try to find its answer tomorrow if i got i will send u
What is the answer
3.5N
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As we know, f=μR
Here,
f→frictional force
μ→coefficient of friction
R→Normal reaction
In inclined plane, R=mgcos∅
So, f = μR
f = (0.7)mgcos30°
f = (0.7)(10)(½)
f = (0.7)(5)
f = 3.5N
Your answer is
Here,
f→frictional force
μ→coefficient of friction
R→Normal reaction
In inclined plane, R=mgcos∅
So, f = μR
f = (0.7)mgcos30°
f = (0.7)(10)(½)
f = (0.7)(5)
f = 3.5N
Your answer is
could u explain it in more detail
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