How Much Force is Needed to Raise a Block on a Frictional Wedge?

[amal]

Homework Statement

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The coefficient of static friction $μ_s$ between the 100kg block and wedge is $0.20$. Determine the magnitude of force $P$ required to raise the block if the rollers are frictionless and wedge massless. You may take $g=9.8m/s^2$

Homework Equations

$f=μ_s*N$

The Attempt at a Solution

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The block will move whenever there is relative motion between block and wedge.

Now,

$N_1+Psinθ=Wcosθ$

$N_1=Wcosθ-Psinθ$

$f=μ_sN_1$

For impending motion,

$f=Psinθ$

That will give:

$P={\frac{Wμcosθ}{cosθ+μsinθ}}$

Plugging in the values,

$P=186.03N$



Only that my teacher thinks that the answer is wrong. According to him, P should be 465N and N1 should be 1073N odd. Though he didn't elaborate, he thinks that something is wrong with my resolution of P. I don't understand how and I also don't understand why the normal is greater than the weight of the block.

Please Help.

 

[haruspex]

It looks like you attempted to attach a diagram, but for whatever reason I'm not able to view it. Can you describe the set-up?

 

[amal]

Yes. I'll give it a shot.
There is a wedge with 15 degrees wedge angle standing on rollers. The block placed is cut in such a way that it's bottom edge (the one which touches the wedge) is parallel to incline of the wedge. The block is placed in vertical channel and there are rollers in there too. Aim is to lift the block in the channel. Force P is applied on wedge in horizontal direction, on the straight side. Friction is between only touching surfaces: incline side of wedge and bottom of the block. I have resolved the force P in two components: along the incline side and perpendicular to incline side.
I am also attaching the figures separately.

 

[haruspex]

There are unknown horizontal forces from the rollers on the block, so it would make sense only to resolve forces on the block vertically. Similarly, there are unknown vertical forces on the wedge. I get a slightly higher value for P than your teacher does, but I've no access to a calculator right now, so that could be my arithmetic at fault.

 

[amal]

What I thought was, as force is a free vector, we could move it to the bottom of block or on the incline surface of wedge. The force is not normal to either surfaces. So, in that case, we generally take two components: one along the incline and one normal to it.
Also, in any case, friction is along the incline. So, if we have not resolved force P, we will have resolve friction. How do we do it? Do we take the component in x-direction as fcos(theta) or f/cos(theta)?

 

[haruspex]

What I thought was, as force is a free vector, we could move it to the bottom of block or on the incline surface of wedge. The force is not normal to either surfaces. So, in that case, we generally take two components: one along the incline and one normal to it.

Yes, but you have ignored the forces from the rollers. Because you resolved along the incline these have a component contribution.

Also, in any case, friction is along the incline. So, if we have not resolved force P, we will have resolve friction. How do we do it? Do we take the component in x-direction as fcos(theta) or f/cos(theta)?

Yes, if you resolve horizontally on the wedge and vertically on the block then the normal force and friction have to be resolved in those directions. It's no harder than resolving gravity as you did.

 

[amal]

I didn't follow you on the last post. Could You please give me some maths?

 

[haruspex]

The rollers either side of the block exert unknown horizontal forces and an unknown torque, so only consider vertical forces on the block. If the normal force between block and wedge is N, frictional force is F, and angle of wedge is theta, what are the vertical forces on the block?
For the same reason, only consider horizontal forces on the wedge. What are those forces?
In equilibrium, what equations can you write?