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1 answer:
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Answer:
250 N
433 N
Explanation:
N = Normal force by the surface of the inclined plane
W = Weight of the block = 500 N
f = static frictional force acting on the block
Parallel to incline, force equation is given as
f = W Sin30
f = (500) Sin30
f = 250 N
Perpendicular to incline force equation is given
N = W Cos30
N = (500) Cos30
N = 433 N
Vector u :
u = 6 i - 3 j
The magnitude of vector u :
| u | =
Answer:
The magnitude of vector u is 3√5.
u = 6 i - 3 j
The magnitude of vector u :
| u | =
Answer:
The magnitude of vector u is 3√5.