Question

5. A block of mass 10 kg is kept on a rough inclined plane as shown in the figure. A force of 3
N is applied on the block. The coefficient of static friction between the plane and the block is
0.6. What should be the minimum value of force P. such that the block doesnot move
downward? (take g=10 ms.​

Answer 1

I am assuming that 'P' is the angle of the incline

The picture above must be the diagram in question

Answer:

approx 51 degrees

Explanation:

First ,write down the given data

• Mass = 10 kg
• coefficient of friction = 0.6
• The force applied = 3 N

To find the minimum force required,

we need to find the force pulling it downwards.

We know that the component of weight (not mass) along the incline is

$mg \cos(x)$

where m is mass

g is accelaration due to gravity

x is the angle of the incline

To find the angle x we need to find the net force.

And as given in the question the net force should be 0

The formule for Net force is

Mgcos(x) - [F(friction) +Force applied] =0

Using the data given above...

we must apply all the values into the formule

That is

$(10 \times 10 \times \cos(x)) - ((0.6 \times 10 \times 10) + 3)$

100 cos(x) - 63 =0

100 cos(x) = 63

cos x = 0.63

cos inverse of 0.63

which is approx 50 degrees