# Mass on a friction less incline. What force is needed to stop the block from sliding.

• danpiz23
In summary, for a block with mass 10kg resting on a smooth, frictionless ramp inclined at 45 degrees, the magnitude of the force required to prevent the block from sliding down the ramp is equal to the weight of the block in Newtons. This is due to the fact that the net force in both the horizontal and vertical directions must be zero for the block to remain stationary. Therefore, the parallel force to the ground must be equal to the weight of the block. This will always be the case for a block on a frictionless ramp.
danpiz23

## Homework Statement

A block with mass 10kg rests on a smooth, friction less ramp that is inclined at an angle of 45 degrees with the ground. How much force must be applied in the direction parallel to the ground to prevent the block from sliding down the ramp?

IT TURNS OUT THAT THE MAGNITUDE OF THE FORCE THAT MUST BE APPLIED IN THE DIRECTION PARALLEL TO THE GROUND IN ORDER TO PREVENT THE BLOCK FROM SLIDING DOWN THE RAMP IS EQUAL TO THE WEIGHT OF THE BLOCK (IN NEWTONS). Explain why this is the care. Will this always be the case for a block on a friction less ramp?

## Homework Equations

I found the answer (please check) as 69.3N, the teacher gave us a follow up page saying the force should be equal to the weight of the block in Newtons..Did I do this wrong?? The follow up page also asks to explain why the parallel force to the ground is equal to the weight of the block in Newtons. I don't get that.

## The Attempt at a Solution

Weight = 10.9.8 = 98N
W vector <0,-98>
Vector u + Vector V = Vector W
F=-u
magnitude of F= magnitude of U
98 sin 45 = 69.3 N
A force of 69.3 N parallel to the plane will keep the weight from sliding.

## The Attempt at a Solution

note the force is applied in a direction parallel to the ground and you must also account for the reaction force, perpendicular to the ramp surface

have you drawn and FBD with all forces?

I drew the fbd with the block sitting on the origin. 3 forces acting on the block Gravity/Normal force and F app.

what am i missing?

lets use your notation, so there are three forces, let's call their magnitudes as follows
W=mg the weight of the mass (vertical)
N the reaction force (unknown, at 45degrees)
F applied force (unknown, horizontal)

in vector notation we have:
W = <mg,0>
N = N<cos(45),sin(45)>
F = <0,-F>

now do the force balance

Also leave numbers out for as long as possible, only substitute in at the end - it generally makes things easier to understand

Last edited:

So fnet=w-n-f?

This is not sure where I'm headed with the equations you gave. I understand those are the forces acting on the mass, but I am looking for the force that keeps the block from sliding. Which the teacher said should be equal to the mass in N of the weight. I so not understand that concept.

danpiz23 said:
So fnet=w-n-f?

This is not sure where I'm headed with the equations you gave. I understand those are the forces acting on the mass, but I am looking for the force that keeps the block from sliding. Which the teacher said should be equal to the mass in N of the weight. I so not understand that concept.

I don't understand your equation, what is fnet?

If the block is stationary the net force is zero in both the horizontal and vertical directions

go back to post #4
- use the sum of vertical forces =0, to find N in terms of W
- then use the sum of horizontal forces = 0 to find F in terms of N

this will give the result you require

## 1. What is a frictionless incline?

A frictionless incline is a smooth, sloping surface where there is no resistance or friction between the object and the surface. This allows the object to slide without any external forces acting on it.

## 2. How is the force needed to stop a block on a frictionless incline calculated?

The force needed to stop a block on a frictionless incline is calculated using the equation F = ma, where F is the force, m is the mass of the block, and a is the acceleration due to gravity. This force is equal to the weight of the block, which is given by the equation F = mg, where g is the acceleration due to gravity (9.8 m/s²).

## 3. Can the angle of the incline affect the force needed to stop the block from sliding?

Yes, the angle of the incline can affect the force needed to stop the block from sliding. The steeper the incline, the greater the force needed to stop the block. This is because the component of the weight of the block acting parallel to the incline increases as the angle increases.

## 4. Is the force needed to stop a block on a frictionless incline constant?

Yes, the force needed to stop a block on a frictionless incline is constant. This is because the weight of the block and the acceleration due to gravity are constant, and the angle of the incline does not change once the block is placed on it.

## 5. How does the mass of the block affect the force needed to stop it from sliding on a frictionless incline?

The mass of the block does not affect the force needed to stop it from sliding on a frictionless incline. This is because the force needed to stop the block is only dependent on its weight and the acceleration due to gravity, which are independent of the block's mass.

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