My significant other told me that you can not exert pressure greater than your weight (without jumping), but was just using ‘common sense’ as his base of knowledge.
While I’m not saying he’s wrong, I’m also not sure he’s right. Is there anyone out there that can confirm/deny this?
As background, here’s the situation under discussion. We ride two up on a motorcycle. I was concerned about how much pressure I was exerting on the foot peg while mounting swing-leg over a lot of gear (large saddle bags and a tail bag). I have a very short inseam so it takes quite a bit of ‘lift’ for me to get my short leg up and over, even starting on the peg. He said “Don’t worry; you can’t exert any more downward pressure than that equal to your body weight.” I think he was just trying to reassure me, but now we are both curious. He suggested I ask on the 'dope, because I’m always telling him how smart ya’ll are.
I don’t have a physics background, but I do know that while I only weigh about 130-140, I can leg press around 200-250. Leg presses involve exerting pressure. I think your boyfriend is very sweet for trying to reassure you, but the way he worded it is wrong.
If you stand still then the force you exert will be your mass. If you accelerate your body in a downward motion then the force will increase beyond your mass.
I believe HillKat was asking if you could exert more pressure downward than you weigh without resorting to external bracing or other mechanical advantage.
Sure, you can put your bathroom scale in a doorway and push against the top of the doorjamb to increase your “weight” on the scale, but that’s not what she’s asking.
Short answer, briefly and slightly.
Take the same bathroom scale used in the previous sample and stand on it going through the same motions you would mounting the bike. Note that you’ll have to have a spring type scale, preferably one that records the max weight. Electronic scales won’t have the response time needed. I’ll WAG that you might add <5% to the max reading if you’re moving very quickly. That’s from accelerating your leg and foot from standing still to high enough to seing over the bike. There will be a comparable decrease in weight when your leg again decelerates to 0. The additional weight will be very slight as the rate of acceleration is very slight.
Note that this is very simplified as a lot of the forces generated are vertical as well.
Dynamic loads are generally accepted to be 1.5 to 2 times the equivalent static load. Try jumping on the hypothetical scale runner pat talks about. I guarantee it’ll read higher than 140lbs (assuming the scale can read in real time). IIRC, the bicycle transmission group in my senior design class used a factor of 1.8 when designing their pedals, crankshaft, and transmission.
As a practical matter, the footpegs are designed with pretty large safety factors, and are probably designed for at least 250 pounds. You’re not going to break it.
You could also take a home scale (spring again not digital) and place it on a chair.
Have your SO watch the dial as you step from the floor up onto the scale in a quick motion. See what the max read is on the scale while doing so.
Well, I understand that this is the rate of acceleration of a free falling object, but right now I am comfortably seated at my desk and not in fact starting my descent from the window ledge.
Gravity is pulling on you with such a force that, if your chair was not holding you, you would be accelerating at 9.8 m/s^2. You aren’t moving simply because the chair is pushing back with enough force to “cancel out” the force of gravity.
And if gravity were to suddenly go away, you would go flying upwards accelerating at about 9.8 m/s^2.
Except that the OP is talking about basically pushing off of something. If I stand on my scale and try to exert force downward, it will only be my weight, buuuuuut if I use the scales to launch myself across the room – or over the piles of laundry lying about on the floor, I am pretty sure the scales will read much higher than 140.
Indeed you would. I left that experiment out of my explanation since the OP specifically excluded jumping.
If you stand on the scale and execute a vigorous high kick there will be a brief increase in apparent weight at the start of the kick and a corresponding decrease at the top of it.
Are we exluding jumping onto a scale or jumping off of a scale or both?
Either would show an increase over her weight.
Mounting the bike using the peg would be similar to jumping off a scale or crouching on a scale and quickly standing up.
A standard practice when riding is to rise up or stand on the pegs when going over rough spots (railroad tracks) or other unavoidable obstacles. This will generate much more force on the pegs than you ever will getting on or off the bike.
A dynamic load can be generated without jumping off the scale, by simply doing a squat quickly.
Neglecting air resistance, during a jump, the readout of the scale you’re jumping off of and the scale you’re landing on should be identical. The rate of change may be different if your legs aren’t bending the same on both scales, but the integral would have the same value.
g. If you are standing still, there’s still 9.8 m/s[sup]2[/sup] to consider, which is your weight.
To the OP: raising your left leg imparts a nontrivial force downward on your right leg, so the peg sees both that AND your body weight. The sum will be way less than 2X your weight, and those pegs will typically be able to handle 500 lbs. You won’t break them unless they’re already damaged.
Wouldnt you slam into the wall at around 1000mph instead?
I am thinking swinging a ball on a string in a horizontal circle around you and then cutting the string…the ball will continue in a straight line in a direction that is tangential to the point where it was in this orbit when the string was cut.
Or can I not equate the string to gravity and you to the ball?
Only O Level physics here, sorry.
