1
$\begingroup$

I was playing around with the block-on-an-incline with friction problem, and imagining Bob having to push a 50 kg block up a 40º slope.

I know friction $f_{s}$ (we will use static friction here, since we are thinking of starting motion) is going to oppose any applied force, including both gravity and the force of Bob pushing his block up the slope. I assumed $\mu = 0.55$

So I went about it this way:

I know $F_N$ (normal force) is going to be $mg\cos(\theta)$ and therefore frictional force $f_f = \mu mg\cos(\theta)$

So far so good, because the force down the slope ($F_{||}$) will be $F_{||} = mg\sin(\theta)$

I plug the numbers in and find that the force that the block would be exerting down the slope, absent friction, is $F_{||} = (50kg)(-9.8m/s^2)(\sin(40º) = -490 N(0.643)= -315.07 N$ (we are going down, so I went negative)

But there is friction, which opposes motion, and it "takes away" from the force down the ramp.

So I did this: $\Sigma F = F_{||}+f_f=-315.07 + f_f$

And $f_f = \mu mg\cos(\theta) = (0.55)(490N)(0.766) = 206.44 N$

So $\Sigma F = F_{||}+f_f=-315.07 + 206.44 N = -108.63 N$ --> so +108.63 N is the minimum force that I have to exert on the block in order to keep it from sliding down the slope. This should be sensible since the force from friction opposing motion is less than what you'd have from gravity all by itself, and it doesn't oppose motion enough to stop the block from sliding. I can even check this by setting up

$F_{||} = f_f \rightarrow mg\sin(\theta) = \mu mg\cos(\theta) \rightarrow \tan(\theta)=\mu$

so $\theta = 28º$ and that's less than the angle I have.

So what, you say? Well, now I turn to how much force Bob needs to push with to get the block moving. I already know how much he needs to hold it in place. Intuitively I would say that is enough to keep the block moving at a constant velocity up the slope (and 0 is a constant too!).

But when I did this calculation adding the gravitational force (-315.07N) to the force from friction (-206.44 N, negative because it is opposing any motion upwards, the direction Bob is pushing) I get a total opposing force of -521.51 N, which says to me that Bob has to push with that much force uphill to get the block moving. In one sense this makes sense, it's harder to get anything to move up a hill than to hold it in place, but I feel like I have made some fundamental error. I would have thought that the force needed to hold the block in place would be equal to what I would get to move the block up at constant velocity. But here I find that just to "balance" the forces (if Bob pushes uphill) he needs a lot more oomph, shall we say.

So I am trying to see if there is an error in my reasoning that it takes more "push" to move something at constant velocity upwards than to hold it in place, even though you are moving at constant velocity in both cases (one just happens to be zero). Because in cases on flat ground if I want to move against friction at constant velocity all I need do is balance the kinetic friction against the force I push with. As long as those are equal, you'll move at constant velocity.

One thought I had was that as you push up, you are opposing frictional force, using energy all the while, and that accounts for the discrepancy; the amount of work (W = Fd) you would be putting in would be greater than holding it in place since the d = 0 in that case, so W = 0 anyway). Also unlike the flat-ground case you are opposing gravity as well.

I thought also that my assumptions about µ might be off; after all static friction is greater (usually) than kinetic, but even if I assume a higher static friction and plug in something lower for kinetic, I get similar results.

Or maybe I did the whole thing correctly and I am overthinking this.

Thanks!

$\endgroup$
1
$\begingroup$

Yes. It will take more to push a block up than to hold it in place. This has to do with static friction. The direction and magnitude of static friction will change depending on how much force is applied.

Consider a slope with horizontal angle $\theta$ and static and kinetic friction coefficient $\mu$. Now, suppose that the block will remain at rest on the ramp without you applying any force. Assume upwards is positive, This implies that $$\Sigma F_\parallel=f_s-F_{g\parallel}=0\qquad\implies\qquad f_s=mg\sin\theta$$

Let's make up some numbers for the sake of clarity. Suppose $mg\sin\theta=10\ \rm N$ which would imply that $f_s=10 \ \rm N$. Also let's suppose that $f_s=10 \rm \ N$ corresponds to the maximum friction. Okay, so far so good.

Now, suppose you apply a $1 \ \rm N$ force $F_A$ upwards on the block. The block won't move. We thus know that $$\Sigma F_\parallel=f'_s+F_A-F_{g\parallel}=0\qquad\implies\qquad f_s'=mg\sin\theta-F_A$$

We can see that friction will now have a smaller magnitude. Using the above values, we now get the static friction is $9\ \rm N$.

Then, let's apply a force of $10 \ \rm N$ upwards on the block. However, the block still won't move. Why? This time, friction will be zero. (because applied force upwards equals gravity, no need for static friction to keep things static).

After that, let's apply a $15 \ \rm N$ force upwards. The block still won't move. But now, friction is acting downwards, and not upwards like before.

$$\Sigma F_\parallel=F_A-F_{g\parallel}-f_s''=0\qquad\implies\qquad f_s''=F_A-mg\sin\theta$$

Using our values, this leaves us with a $5 \ \rm N$ friction force downwards.

By this reasoning, the force required to hold the block in place such that a slight push would accelerate the block) is $20 \ \rm N$. If we were to push the block up at a constant speed, the force we'd apply would be $20 \rm N$.

Therefore, the force needed to hold an object in place is not necessarily the force needed to push an object up at constant speed.

The force required to move your block at a constant speed up the ramp is the same as the force needed to hold it in place only if the the block is on the verge of moving (i.e. friction can't get any larger or change its direction to oppose motion).

$\endgroup$
1
  • 1
    $\begingroup$ Thanks a lot, this does help clarify things! I felt like I was intuitively correct but wanted to be sure that it stood up to scrutiny, and I was sure I had done things correctly, but it didn't "feel" right. $\endgroup$ – Jesse Feb 10 at 19:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.