Calculating the Force on a Sloped Wedge & Scale

[blakeoman]

What is the force that the scale reads (the "weight") as the mass slides down the slo

There is an object on top of a wedge sitting on a scale. When the object is released, the object slides down the slope of the wedge. What is the force that the scale reads (the "weight") as the mass slides down the slope?

There is friction between the object and the wedge and between the wedge and the scale. The static friction between the wedge and the scale is large enough to keep the wedge from moving.

The mass of the object is 1.2 kg.

The mass of the wedge is 7.5 kg.

The angle of the wedge is 21 degrees.

The coefficient of kinetic friction between the object and wedge is 0.11.

Give your answer in Newtons and to three significant digits.

 

[JoshuaR]

What have you tried thus far?

 

[blakeoman]

I set up all the free body diagrams and laid out what i think is all the X and Y forces but now I am stuck.

 

[djeitnstine]

I set up all the free body diagrams and laid out what i think is all the X and Y forces but now I am stuck.

The scale reads $$F_{N}$$ which is the normal force $$F_{N}=mgcos\Theta$$

depending on where the scale is and what it is supposed to be reading.

 

[blakeoman]

The scale reads $$F_{N}$$ which is the normal force $$F_{N}=mgcos\Theta$$

depending on where the scale is and what it is supposed to be reading.

here is the picture from the problem.

 

[djeitnstine]

Scale will only read forces in the y direction, this seems like a trick question. The scale is measuring ALL of the masses. So whether its on incline or not. $$\sum{\vec{F}} = m_{block} \vec{g} + m_{wedge} \vec{g}$$

 

[blakeoman]

Scale will only read forces in the y direction, this seems like a trick question. The scale is measuring ALL of the masses. So whether its on incline or not. $$\sum{\vec{F}} = m_{block} \vec{g} + m_{wedge} \vec{g}$$

Will the scale reading be different if the object is moving compared to if the object is not moving down the wedge?

 

[djeitnstine]

Lets see, the masses remains the same, and acceleration is still g

 

[blakeoman]

I got it!
Fscale= Fwedge + Fblock
Fwedge= G * Mwedge
Fblock= Fnormal + Ffriction= Fnormal + (friction)*Fnormal=Fnormal*(1+(friction))=Mblock*G*cos(theta)*(1+(friction))
so,
G*Mwedge + G*Mblock*cos(theta)(1+(frictionforce))

 

[PhanthomJay]

I got it!
Fscale= Fwedge + Fblock

The scale reads vertical forces only. What do you mean by Fblock?

Fwedge= G * Mwedge

this is the weight of the wedge, correct.

Fblock= Fnormal + Ffriction= Fnormal + (friction)*Fnormal=Fnormal*(1+(friction))=Mblock*G*cos(theta)*(1+(friction))

the normal and friction forces are perpendicular to each other; you can't just add them

so,
G*Mwedge + G*Mblock*cos(theta)(1+(frictionforce))

Since the object is sliding down the wedge, that means it must be accelerating both vertically and horizontally, and the scale must read less than the total weight of the block and wedge. Isolate the top block first, and identify all force components in their correct x and y directions, and apply Newton 2.

 

[marlo23]

I have almost the EXACT same problem...so what's the correct equation??

 

[nasu]

You have 3 forces acting on the big bloc: Fn, Ff and its weight, M*g.
The vertical components of Fn and Ff will add to M*g.
Be careful with the components and the directions of the forces (the friction on the block, not on the small object). You can easily verify your answer.
The answer is OK if it goes to M*g for an angle of 90 degrees and to M*g + m*g for an angle of zero degrees.

 

[JoshuaR]

Let's say you put a block on a flat scale. It measures weight of mg. You exert force exactly parallel to the scale, does the reading increase? How about it you put a second bloc on top of the first? Now the scale reads (M+m)g. How about if you slide one block but not the other, using only a horizontal force? Does the reading change?

Now to the wedge problem. You have two objects of some mass. Independently, what are the forces on the objects? Just gravity down, and the scale up (normal force, equal and opposite). Stack the blocks on top of each other, how does it work? etc.

 

[unscientific]

I know the answer. Draw the vector triangle for the acceleration;

acceleration parallel to the surface of slope is gsintheta, then the downward accleration is (g sin (sin(theta) ).

Drawing the freebody diagram,

Weight - Normal force = Mgsin^2 theta

therefore,

Normal force = Mg - Mgsin^2 theta = Mg (cos^2 theta)

 

[unscientific]

look at the picture, the object on the weighing scale is accelerating parallel to the surface of the wedge at gsintheta, then when you draw the vector diagram for the acceleration, just sine it 1 more time, getting g sin²theta.

Hope this helped, this is the shortest way I can find to solve this question :)

 

[nasu]

look at the picture, the object on the weighing scale is accelerating parallel to the surface of the wedge at gsintheta, then when you draw the vector diagram for the acceleration, just sine it 1 more time, getting g sin²theta.

Hope this helped, this is the shortest way I can find to solve this question :)

This force (m*g*sin(theta)) acts on the small object and not on the big block.
You need the forces acting on the big block and these are the normal and the friction (plus its own weight).

The total vertical force on the big block will be M*g+m*g*cos(theta)^2+miu*m*g*cos(theta)sin(theta)
The second term is the vertical component of the normal force and the third is the vertical component of the friction.