Question

A block weighing W = 10 kN is resting on an inclined plane . Determine its
components normal to and parallel to the inclined plane.
​

Answer 1

Answer:

6.062

Explanation:

Given that,

Weight of block = 10 N

Mass =  

10

10

​  

=1 kg

Coefficient of friction μ=0.7

We know that,

Normal reaction = mg cosθ

 N=mgcosθ

N=Wcosθ

N=10×cos30  

0

 

N=10×  

2

3

​  

 

​  

 

N=5  

3

​  

 

Now, frictional force

 F=0.7×5  

3

​  

 

F=6.062N

Hence, the frictional force is 6.062 N.