Answer:
70.6N
Explanation:
We are given that
Mass of crate=12 kg
[tex]\theta=30^{\circ}[/tex]
Coefficient of static friction=[tex\mu=[/tex]0.40
Horizontal component force is equal to friction force
[tex]Fcos\theta=f=\mu N[/tex]
Force along vertical direction
[tex]Fsin\theta+mg=N[/tex]..(2)
Using equation(2) in equation (1)
[tex]Fcos\theta=\mu(Fsin\theta+mg)[/tex]
[tex]Fcos\theta=\mu Fsin\theta+\mu mg[/tex]
[tex]\mu mg=Fcos\theta-\mu Fsin\theta=F(cos\theta-\mu sin\theta)[/tex]
[tex]F=\frac{\mu mg}{cos\theta-\mu sin\theta}[/tex]
Substitute the values then we get
[tex]F=\frac{2\times 0.40\times 9.8}{cos 30-0.40sin 30}[/tex]
[tex]F=70.6N[/tex]
Hence, the minimum force needed to start the crate moving=70.6N
