Answer:
The magnitude of applied force,parallel to the incline is 575.38 N and parallel to the floor is 605 N.
Explanation:
Given:
Mass of the piano [tex](m)[/tex] = 190 kg
Inclined angle [tex](\theta)[/tex] = 18 degree
Considering gravity, [tex]g[/tex] = 9.8 [tex]ms^-^2[/tex]
And
Using, [tex]sin(18) =0.30[/tex] and [tex]cos(18)=0.95[/tex]
FBD diagram is attached with all the force acting on the floor and and the inclined.
We have to find the magnitude of forces,when the man pushes it parallel to the incline and to the floor.
a.
When the man pushes it parallel to the incline.
Balancing the forces as [tex]\sum F=0[/tex] .
⇒ [tex]F+mgsin(\theta) =0[/tex]
⇒ [tex]F=-mgsin(\theta)[/tex]
⇒ Here it is negative as the force is acting downward.
⇒ Plugging the values of mass [tex](m)[/tex] and angle [tex](\theta)[/tex] .
⇒ [tex]F=190\times 9.8\times sin(18)[/tex]
⇒ [tex]F=575.38[/tex] N
b.
When the force is parallel to the floor.
⇒ [tex]Fcos(\theta)=mgsin(\theta)[/tex]
⇒ [tex]F=\frac{mgsin(\theta)}{cos(\theta)}[/tex]
⇒ Plugging the values.
⇒ [tex]F=\frac{190\times 9.8\times sin(18)}{cos(18)}[/tex]
⇒ [tex]F=605[/tex] N
So,
The magnitude of applied force in inclined direction is 575.38 Newton and parallel to the floor is 605 N.
