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Problem 3: A block of mass m= 15.5 kg rests on an inclined plane with a coefficient of
static friction of us = 0.11 between the block and the plane. The inclined plane is L = 6.8 m long and
it has a height of h = 3.05 m at its tallest point.
A. What angle in degrees does the plane make with respect to the horizontal? B. What is the magnitude of the normal force in newtons that acts on the block? C. What is the component of the force of gravity along the plane, fgx in newtons? D. Write an expression in terms of degrees, the mass, the coefficient of the static friction, and the gravitational constant, for the magnitude of the force due to static friction, just before the block begins to slide.

Respuesta :

onyebuchinnaji

(a) The angle the plane makes with respect to the horizontal is 27⁰.

(b) The magnitude of the normal force acting on the block is 135.3 N.

(c) The component of the force of gravity along the plane, fgx is 68.96 N.

(d) The expression for the net force on the block is mgsinθ - μmgcosθ = 0.

Angle of the plane with respect to the horizontal

sin θ = h/L

where;

  • θ is the angle of inclination
  • h is height of the incline
  • L is length of the plane

sin θ = 3.05/6.8

sin θ = 0.4485

θ = arc tan(0.4485)

θ = 26.7⁰

θ ≈ 27⁰

Normal force on the block

Fₙ = mgcosθ

Fₙ = (15.5)(9.8)cos(27)

Fₙ = 135.3 N

Parallel component of the force on the plane

Fx = mgsinθ

Fx = (15.5)(9.8) sin(27)

Fx = 68.96 N

Net force on the block

∑F = 0

Fx - Ff = 0

mgsinθ - μmgcosθ = 0

Thus, we can conclude the following;

  • the angle the plane makes with respect to the horizontal is 27⁰,
  • the magnitude of the normal force acting on the block is 135.3 N,
  • the component of the force of gravity along the plane, fgx is 68.96 N, and
  • the expression for the net force on the block is mgsinθ - μmgcosθ = 0.

Learn more about static friction here: https://brainly.com/question/13680415

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