Answer to Question #96616 in Mechanics | Relativity for Fivekay

Question

mail A horizontal force of 45 newtons applied to a mass of 9 kilograms is just coefficient to move it if the body is now pulled at an angle of 50 to the horizontal, find the force required to move the body over the horizontal surface

Answer 1

We have:

$$N=mg-F\sin{50}$$

$$ma=F\cos{50}-\mu (mg-F\sin{50})=0$$

The friction coefficient:

$$\mu=\frac{F\cos{50}}{mg-F\sin{50}}$$

Now:

$$ma=F'-\mu mg=0$$

$$F'=mg\frac{F\cos{50}}{mg-F\sin{50}}$$

$$F'=(9)(9.8)\frac{45\cos{50}}{(9)(9.8)-45\sin{50}}=47.5\ N$$