Calculating the Frictional Force on a 15kg Mass at 30°

[Dazz4C]

Homework Statement

Given an object of mass 15kg sits on a slide, at an angle of 30 degrees to the horizontal, what is the magnitude of the frictional force acting on the box if the box does not slide down?

Homework Equations

So, I know this is a scalar vector; so I'll be using the Dot Product.

The Attempt at a Solution

I've plotted it as a vector (attached below)

I've solved for 'Length x' using $$Sin 30=x/15$$ which turns out to be 7.5

From this, I used the dot product where ||a|| * ||b||*Cos$$\theta$$ = Friction

Let:
||a|| = 15
||b|| = 7.5
$$\theta$$ = 30

Giving the friction as approx. 97.42...not sure if this is the right process though.

 

[Sethric]

It would be easier to think of this in forces. The frictional force would be equivalent to the component of the force due to gravity that would point down the ramp, since they must cancel. Break gravity down to find that. Are you required to use a dot product here?

 

[Dazz4C]

It would be easier to think of this in forces. The frictional force would be equivalent to the component of the force due to gravity that would point down the ramp, since they must cancel. Break gravity down to find that. Are you required to use a dot product here?

The question is off a worksheet which had required me to use Dot/Cross Product all the way through it, so I've kind of assumed that I'd have to use it for this question.

 

[Sethric]

Hmm, weird. OK. Then you will want to find the projection of the force due to gravity on a vector that points down the ramp. That will require a dot product. Recall projections?

 

[mege]

Your friction vector should be (using your diagram's orientation) at 150 degrees.
OR - think about it with the box and ramp flat - friction is applied to the left, and gravity is applied 30 degrees below the horizon to the right (thinking about the problem in this form may make some of the math easy rather than dealing with a bunch of sin30 and cos30 you get some sin0 and cos0).

You don't neccessarilly need to use dot product if you vectorize everything using cos and sin and your magnitude. You should have a few equivalencies with your vectors and be able to solve for the missing links in each.

 

[tiny-tim]

Hi Dazz4C!

… So, I know this is a scalar vector; so I'll be using the Dot Product.

There's no such thing as a scalar vector … did you mean something different?

The question is off a worksheet which had required me to use Dot/Cross Product all the way through it, so I've kind of assumed that I'd have to use it for this question.

Definitely dot product …

you won't be needing the cross product until you get to rotational motion.

Every component of force is the dot product of the force with the unit vector in that direction.

 

[Dazz4C]

Righto, here's my new attempt:

We know:
Weight of Box (Magnitude a) of 15kg
Gravity's Force (Magnitude b) which acts down on the box 9.8ms^-2
Friction is unknown
Angle between is 30
Friction is a vector quantity, so we use cross product ||a||*||b||*Sin Theta

So:

$$||a||\times||b||\times Sin\Theta$$
$$15kg\times9.8ms^{-2}\times Sin30$$
$$=73.5kgms^{-2} = 73.5N$$

Is that right?

 

[Sethric]

Umm, that is the correct answer, but not the correct method. You shouldn't need the cross product. You defined mass as a vector, and then gravity as a vector, and then said they point in different directions. That is incorrect. Mass is a scalar, force from gravity is a vector. You want to find the projection of the force vector due to gravity on the ramp. (The reason for this is that you are assuming that friction is cancelling out the force that gravity "projects" down the ramp.) So just dot a unit vector in the direction of the ramp with the weight vector. You will end up with the exact same terms, except the 15*9.8 will be wrapped up in the magnitude of just one of the vectors.

 

[Dazz4C]

Umm, that is the correct answer, but not the correct method. You shouldn't need the cross product. You defined mass as a vector, and then gravity as a vector, and then said they point in different directions. That is incorrect. Mass is a scalar, force from gravity is a vector. You want to find the projection of the force vector due to gravity on the ramp. (The reason for this is that you are assuming that friction is cancelling out the force that gravity "projects" down the ramp.) So just dot a unit vector in the direction of the ramp with the weight vector. You will end up with the exact same terms, except the 15*9.8 will be wrapped up in the magnitude of just one of the vectors.

Well, if we know that the frictional force and force from gravity cancels (equals zero) then we can say that Friction Force=Gravity Force. Since we also know that Gravity due to acceleration = mass*gravity*Sin theta (Vector Quantity), then Friction Force is also equal to mass*gravity*Sin theta.

Hence, it ends up being a cross product...I don't see why it's an incorrect method. The magnitudes are already given.

 

[Sethric]

The reason you get a vector is that you are looking at a projection along a vector. In this case, you would be multiplying your scalar projection by a unit vector in that direction. Just because your terms matched the magnitude of a cross product, does not mean you have a cross product. You won't see cross products in mechanics until torque. The reason your terms matched is because sin(30) = cos(60). The question even asks for magnitude, a scalar. You wouldn't expect to find a vector.

 

[Dazz4C]

The reason you get a vector is that you are looking at a projection along a vector. In this case, you would be multiplying your scalar projection by a unit vector in that direction. Just because your terms matched the magnitude of a cross product, does not mean you have a cross product. You won't see cross products in mechanics until torque. The reason your terms matched is because sin(30) = cos(60). The question even asks for magnitude, a scalar. You wouldn't expect to find a vector.

Ahh...okay I understand what you mean.

So would it be somewhat like this diagram? (attached)

 

[Sethric]

No, mass is not a vector. Your two vectors will be a unit vector pointing in the direction of the ramp and a weight vector pointing down. The weight vector will have a magnitude of (9.8)(15) = 147 (Newtons, if it matters).

 

[Dazz4C]

Uhm...what about now?

 

[Sethric]

Correct.