Question

A force of 300 N, acts downwards on a 30 kg block at 40° to the horizontal as shown in the diagram below. The block accelerates to the right along a rough horizontal surface.

Answer 1

Given:

A force of 300 N, acts downwards on a 30 kg block at 40° to the horizontal as shown in the diagram.

To find:

• Magnitude and direction of the component of the 300 N force which accelerates the block horizontally.

• Friction when acceleration of block is 2 m/s².

• Vertical force exerted by block on floor.

Calculation:

1st question:

The required component of Force is:

$f = 300 \times \cos( {40}^{ \circ} )$

$= > f = 300 \times 0.76$

$\boxed{ = > f = 229.31 \: N }$

This component is acting in the +X axis.

2nd Question:

$f - friction = ma$

$= > 229.3 - friction = 30 \times 2$

$= > 229.3 - friction = 60$

$= > friction = 229.3 - 60$

$\boxed{ = > friction = 169.3 \: N }$

Question 3:

Vertical force exerted will be sum of Weight of block and the component of applied force:

$= mg + 300 \sin( {40}^{ \circ} )$

$=(30 \times 10)+ 300 \sin( {40}^{ \circ} )$

$=300+ 300 \sin( {40}^{ \circ} )$

$=300 \{1 + \sin( {40}^{ \circ} ) \}$

$= 300 \times 1.64$

$\boxed{ = 492.8 \: N }$

Hope It Helps.