The weight of a block on the inclined plane is 500 N and the angle of incline is 30 degrees. What is the magnitude of the force
of friction for static equilibrium? What is the normal force of the surface pushing up on the block?
1 answer:
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
You might be interested in
Answer:
Explanation:
The cross product of two vectors is given by
Where, θ be the angle between the two vectors and \widehat{n} be the unit vector along the direction of cross product of two vectors.
Here, K x i = - j
As K is the unit vector along Z axis, i is the unit vector along X axis and j be the unit vector along axis.
The direction of cross product of two vectors is given by the right hand palm rule.
So, k x i = j
j x i = - k
- j x k = - i
i x i = 0