A 10 kg weight is on an inclined plane 10 ft long and 6 ft high. If there is no friction, what force parallel to the plane will prevent the body from sliding?

Answer 1

Given: 10 kg, 10 ft and 6 ft.

Solution:
Find the force parallel to the plane

Sin A = opposite/hypotenuse
Sin A = 6ft/10ft
= 0.6 ft.

Find the A of the sin.

A = sin^-1(6ft/10ft)
A = 36.87°

Find f1 and f2:
Note: m (mass)
g (gravity pulls) = 9.81 m/s^2

F1= sin (teta) x mg
= sin 36.87° x 10 kg(9.81m/s^2)
F1 = 58.86 N prevent the body from sliding

F2 = cos (teta) x mg
F2 = cos 36.87° x 10 kg(9.81m/s^2)
F2 = 78.48 N vertical mass

Answer 2

(6/10) 10 g = 6 g = 6*9.81 = 58.86 Newtons