Answer:
[tex]F=mg(sin(\theta )-0.25 cos(\theta ))[/tex]
Explanation:
The free body diagram of the block on the slide is shown in the below figure
Since the block is in equilibrium we apply equations of statics to compute the necessary unknown forces
N is the reaction force between the block and the slide
For equilibrium along x-axis we have
[tex]\sum F_{x}=0\\\\mgsin(\theta )-\mu N-F=0\\\therefore F=mgsin(\theta)-\mu N......(\alpha )\\Similarly\\\sum F_{y}=0\\\\N-mgcos(\theta )=0\\\therefore N=mgcos(\theta ).......(\beta )\\\\[/tex]
Using value of N from equation β in α we get value of force as
[tex]F=mg(sin(\theta )-\mu cos(\theta ))[/tex]
Applying values we get
[tex]F=mg(sin(\theta )-0.25 cos(\theta ))[/tex]
Answer:
[tex]F = \frac{mgsin\theta}{\mu} - mgcos\theta[/tex]
Explanation:
As we know that the force applied is perpendicular to inclined plane
So here the normal force on the block is given as
[tex]N = F + mgcos\theta[/tex]
now in order to stop the block by the perpendicular force we can say that the friction force is sufficiently large to balance the force of gravity
So here we can say
[tex]F_f = mgsin\theta[/tex]
[tex]\mu N = mgsin\theta[/tex]
[tex]\mu(F + mg cos\theta) = mg sin\theta[/tex]
now we have
[tex]F = \frac{mgsin\theta}{\mu} - mgcos\theta[/tex]
