A horizontal force keeps a 2.0 kg box in equilibrium on a 60-degree incline; what is the magnitude of the horizontal force, and what is the normal force?

Question

Please show this using an x-axis that is parallel to the incline, and please show again using an x-axis that is horizontal.

 

Thank you.

 

Answer 1

assuming no friction, the incline component of the horizontal force = 2kg(10m/s²)sin 60 degrees = 10√3 N

 

therefore,   cos 60 degrees = 10 √3 / (horizontal force)

 

and horizontal force = 10√3 (2) = 20√3 N

 

 

therefore sin 60 degrees = normal force / 20√3)

 

and normal force = 30N 

Answer 2

The weight, vertical force, is 2 X 10 = 20N (10 = gravity acceleration)

Component parallel to the incline, is 20 sin 60 = 20(√3/2) = 10√3

then the horizontal force will have two components, one parallel to the incline and other normal to the incline. 

The parallel needs to be 10√3

and the horizontal force needs to be

(10√3)/ cos 60 = 20√3

For the normal force we have that Tan 60 = FH/Fn

20√3/tan60 = Fn

20√3/√3 = 20 N

 

 

 

 

 

 

 

 

 

 

Answer 3

Since box is not moving, the forces are balanced. There are three forces: gravity, normal force, and horizontal force F.

 

1) Let x-axis be parallel to the incline. In this case, component of force F along this axis is F*cos(60). Component of the gravity force along the same axis is -m*g*sin(60) (negative sign indicates opposite direction). The sum of two forces must be zero. That is,

F*cos(60)-mg*sin(60)=0. From this you obtain:

 

F=mg*tan(60)=2.0*9.8*√3≈34N

 

Normal force has zero component along x-axis. Along the vertical axis we have:

 

N_y=N; F_y=-F*sin(60); mg_y=-mg*cos(60)

 

Again, sum of all three shall be zero. Thus we obtain:

 

N-F*sin(60)-mg*cos(60)=0. Plugging in expression for F, F=mg*tan(60), we obtain, after some algebra:

 

N=mg/cos(60); N=19.6 N/(½)≈39N

 

2) If x-axis is horizontal, then two equations will look as follows:

 

F-N*sin(60)=0

N*cos(60)-mg=0

 

Solving for N yields:

 

N=mg/cos(60)

 

Using the expression for N we again obtain for F:

 

F=mg*tan(60)

 

Numerical calculations are the same, of course.

 

Notice that normal force is greater than gravitational one in this case! This is because there is external force F, which creates additional weight and increases the normal force, which must balance the object's weight. So the weight of an object can be less than gravitational force on it and can be greater than the force of gravity.