Question

a block(8.5kg) is fixed in aa frictionless incline plane by a rope. the angle θ= 30. what is thnormal force acting on the block.

Answer 1

Answer on Question #44606, Physics, Mechanics | Kinematics | Dynamics

a block(8.5kg) is fixed in aa frictionless incline plane by a rope. the angle $\theta = 30$ . what is the normal force acting on the block.

Solution.

From $1^{\mathrm{st}}$ Newton's law:

$$\overline {{m g}} + \vec {N} + \vec {F} = 0$$

Where $N$ is the normal force acting on the block.

In projections:

OX: $mg \cdot \sin(\theta) = F$

OY: $mg \cdot \cos(\theta) = N$

So: $N = mg \cdot \cos(\theta)$

Numerically:

$$N = 8. 5 k g \cdot 9. 8 \frac {m}{s ^ {2}} \cdot \cos (3 0 {}^ {\circ}) \approx 7 2. 1 4 N$$

Answer: $N \approx 72.14$ N

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