a) mg cos(24)-N=0
b) mg sin(24)=ma
c) Since there is no original figure, in my coordinates a.y=0.
d) According to a): N=331.25 N
Answer:
a) Along X direction since it can not move so Net force in x direction must be ZERO
[tex]F_{net} = F_n - mgcos\theta[/tex]
since a = 0
[tex]F_n - mgcos\theta = 0[/tex]
b)For Y direction net force is only due to component of the weight of object along the inclined plane
[tex]mgsin\theta = F_{net}[/tex]
[tex]m a_y = mgsin\theta[/tex]
c) for finding acceleration in Y direction we can use above equation
[tex]m a_y = mg sin\theta[/tex]
divide both sides by mass
[tex]a_y = g sin\theta[/tex]
[tex]a_y = 9.8 sin24 = 3.98 m/s^2[/tex]
d) For finding Normal force we can use the expression of force in X direction
[tex]0 = F_n - mg cos\theta[/tex]
[tex]0 = F_n - (37\times 9.8\times cos24)[/tex]
[tex]0 = F_n - 331.25[/tex]
[tex]F_n = 331.25 N[/tex]
