# A block of mass 11 kg is at rest on a rough plane inclined at 30° to the horizontal. A force acts on the block in a direction up the plane parallel to a line of greatest slope. When the magnitude of the force is 2X N the block is on the point of sliding down the plane, and when the magnitude of the force is 9X N the block is on the point of sliding up the plane. Find the value of X.

My work so far isn't much, just one equation: 9X - F = 110sin30

I know that I have to get another equation and cancel them out the same way you'd do simultaneous equations, but I'm not so sure how to get the second one.

Thanks for the help!

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## forces on the block are

... friction (F) ... opposing motion up or down
... gravity (G) ... downward along plane
... X force ... upward along plane

gravitational force perpendicular to the plane is part of the frictional force

G = F + 2X

G + F = 9X

you already have G as 110 sin(30º)

eliminate F , and you've got it

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## Well, here's a humorous take on your problem:

Ah, the block on a rough inclined plane - a classic tale of forces and friction! Let's slide into solving this together, shall we?

First, let's tackle the equation you already have: 9X - F = 110sin30. It looks like you're headed in the right direction, so I won't clown around with it.

Now, to obtain the second equation, we need to consider the point at which the block is on the verge of sliding down. In this case, the force acting up the plane just balances out the force of gravity acting down the plane, with some friction thrown into the mix.

So, let's clown around with some physics equations! The force of gravity acting down the plane can be calculated as m*g*sinθ, where m is the mass of the block and θ is the angle of inclination.

In this scenario, the force acting up the plane is just enough to counterbalance the force of gravity, so we can set up the equation: F = m*g*sinθ.

But wait, we need X in this equation! Fear not, as we can use the given information. When the magnitude of the force is 2X N, the block is on the verge of sliding down the plane. So we can substitute 2X for F in our equation: 2X = m*g*sinθ.

Now that we have our second equation, we can put on a juggling act and solve these simultaneous equations to find the value of X. Simply combine the two equations and eliminate F. Clown Bot out!

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## To find the second equation, let's consider the equilibrium conditions for the block on the inclined plane.

When the block is on the point of sliding down the plane, the force parallel to the incline just balances the component of the weight acting down the incline. This condition is given by:

F = (m * g * sinθ)

where F is the force parallel to the incline, m is the mass of the block, g is the acceleration due to gravity, and θ is the angle of inclination.

So, in this case, F = 2X N and θ = 30°, and the equation becomes:

2X = (11 * 9.8 * sin30)

Simplifying,

2X = (11 * 9.8 * 0.5)
2X = 53.9

Now, let's find the second equation for when the block is on the point of sliding up the plane.

When the block is on the point of sliding up the plane, the force parallel to the incline just balances the component of the weight acting down the incline. This condition is given by:

F = (m * g * sinθ)

So, in this case, F = 9X N and θ = 30°, and the equation becomes:

9X = (11 * 9.8 * sin30)

Simplifying,

9X = (11 * 9.8 * 0.5)
9X = 53.9

Now we have two equations:

2X = 53.9
9X = 53.9

To find the value of X, we can solve these two equations simultaneously.

Dividing the first equation by 2 and the second equation by 9:

X = 26.95
X = 5.99

Since we're looking for the value of X, we choose the smaller value.

Therefore, the value of X is approximately 5.99.

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## To solve for the value of X, you need to set up another equation using the given information.

Let's consider the forces acting on the block along the inclined plane. There are two forces acting in that direction: the gravitational force mg (where g is the acceleration due to gravity) acting downwards and the force parallel to the inclined plane, which is 2X N when the block is on the point of sliding down and 9X N when the block is on the point of sliding up.

When the block is on the point of sliding down the plane, the force parallel to the plane (2X N) is equal to the maximum frictional force opposing its motion. The frictional force can be calculated as the product of the coefficient of friction (μ) and the normal force (N) exerted by the block on the plane. The normal force is equal to the component of gravitational force perpendicular to the plane, which is mg cosθ, where θ is the inclination angle.

Therefore, the maximum frictional force is μmg cosθ. However, since the block is just on the point of sliding, the frictional force is at its maximum value and can be represented as μmg cosθ = 2X N.

Similarly, when the block is on the point of sliding up the plane, the force parallel to the plane (9X N) is equal to the sum of the gravitational force acting downwards and the frictional force acting upwards. Hence, we have 𝑚gsinθ + μ𝑚gcosθ = 9X N.

Now, we can substitute the value of μmg cosθ from the first equation into the second equation:

𝑚gsinθ + 2X + 𝑚gcosθ = 9X

Simplifying the equation:

𝑚g(sinθ + cosθ) + 2X = 9X

Finally, you have two equations:

1. 9X - μmg cosθ = 110sinθ
2. 𝑚g(sinθ + cosθ) + 2X = 9X

Now you can solve these two equations simultaneously for the value of X.