Answer:
253 N
Explanation:
The forces acting on the box in the direction along the ramp are two:
- The component of the weight in the direction parallel to the ramp, [tex]mg sin \theta[/tex], acting down along the ramp
- The force of static friction, [tex]F_f[/tex], acting up along the ramp
The equation of motion along this direction is:
[tex]mg sin \theta - F_f= ma[/tex]
where:
m = 55.0 kg is the mass of the box
[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity
[tex]\theta=28^{\circ}[/tex] is the angle at the base of the ramp
a is the acceleration of the box
However in this proble, the box is at rest, so the acceleration is zero:
[tex]a=0[/tex]
Therefore, the equation becomes:
[tex]mg sin \theta -F_f[/tex]
From which we find the static frictional force:
[tex]F_f=mg sin \theta = (55.0)(9.8)(sin 28.0^{\circ})=253 N[/tex]
The static frictional force which holds the box in place will be:
"253 N".
According to the question,
Mass of box, m = 55.0 kg
Acceleration due to gravity, g = 9.8 m/s²
Angle, θ = 28°
By using equation of motion,
→ mg Sinθ - [tex]F_f[/tex] = ma
When the box is at rest, the acceleration "a" be "zero (0)".
then,
→ mg Sinθ - [tex]F_f[/tex] = 0
hence,
The static frictional force be:
[tex]F_f[/tex] = mg Sinθ
By substituting the values,
   = 55.0 × 9.8 × Sin 28.0°
   = 253 N
Thus the above answer is correct.
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