Question

Na figura, um bloco de massa igual a 5Kg é abandonado em repouso em um plano inclinado. Supondo que os coeficientes de atrito estático e dinâmico sejam idênticos e valem µe = µd = 0,90 e sendo θ=36,8°, determine: a) a intensidade da componente P em x b) a intensidade da força Normal c) a intensidade da força de atrito d) a aceleração do bloco (justifique sua resposta)

Answer 1

The figure mentioned on the question is in the attachment.

Answer: a) $P_{x}$ = - 38.35N

b) $F_{N}$ = 30.5 N

c) $F_{f}$ = 27.45 N

d) a = - 13.16 m/s²

Explanation: A block on an inclined plane has 3 forces acting on it:

• Force due to gravity $F_{g}$ = m.g;
• Normal Force due to the plane;
• Force of Friction $F_{f}$ = µ.N;

Since the plane is inclined, Normal Force is equal the y-component of the force due to gravity and Force of friction and the x-component of the force due to gravity are opposite forces.

The second attachment ilustrate the forces acting on the block.

Calculating:

A) The magnitude of the x-component of Force due to gravity:

According to the second image:

$P_{x}$ = P.sinθ

$P_{x}$ = 5.9.8.sin(36.8)

$P_{x}$ = - 38.35 N

B) $F_{N}$ = $P_{y}$ = m.g.cosθ

$F_{N}$ = 5.9.8.cos(36.8)

$F_{N}$ = 30.5 N

C) $F_{f}$ = 0.9.30.5

$F_{f}$ = 27.45 N

D) For the acceleration, use Newton's Law: $F_{R}$ = m . a

If there is movement, it is only on x-axis, so the net force is:

$P_{x}$ - $F_{f}$  = m.a

- 38.35 - 27.45 = 5a

a = - 13.16 m/s²

The value of acceleration shows there is no movement on the x-axis due to the friction.