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An overweight acrobat, "weighing" in at 115 kg, wants to perform a single handstand. He tries to cheat by resting one foot against a smooth frictionless vertical wall. The horizontal force there is 130 N. What is the magnitude of the force exerted by the floor on his hand?
Answer in Newtons

 

Please explain this problem thuroughly. 

 May 15, 2019
 #1
avatar+116 
+4

Visualizing this problem can go a few different ways, but I'm just gonna keep it in simple geometry.

 

Like, really simple geometry lmao.

 

To explain the numbers, the horizontal 130 N is given to us. The downward force, or weight of the acrobat, is \(115 * 9.8 = 1127\).

 

Now, what is the magnitude of the force exerted by the floor on his hand? Well, it definitely isn't a normal force with two perpendicular forces. Oo! Then we can just use the resultant of the two forces. In other words, the magnitude of the force exerted by the floor on his hand is the hypotenuse of the above triangle in the drawing.

\(\sqrt{1127^2+130^2} = 1134.473N\)

.
 May 15, 2019
 #3
avatar+14 
0

This result is incorrect. The vertical and horizontal forces are separate and do not add in that manner. Imagine standing next to a wall and start pushing perfectly horizontally on that wall, do you feel heavier? If you push hard enough on a wall will your legs eventually give out from too much force? I hope not. The vertical normal force on the hand is only the 1127 N. The 130N that is applied horizontally would result in a torque force on the hand, and an increase in frictional force on the hand, not actual pushing forces. What you were calculating is the magnitude of the total force on the acrobat, not the force on the hand.

Benn.A  May 15, 2019
edited by Benn.A  May 15, 2019
 #4
avatar+102447 
+1

Thanks for your answers Anthrax and Ben.  laugh

 

But...

Ben, it would have sounded an aweful lot nicer if you had started with something like:

'Antrax, I do not think your answer is right because .......'

If you had actually asked the question, which you didn't, then I'd also suggest that you started with 'Thanks'

 

Opening with

'You are wrong', 

or   'Your result is incorrect' 

or similar statements is a great way to get our esteemed answerers off side, and there is nothing acheived by doing that.  wink

Melody  May 15, 2019
 #5
avatar+14 
+2

Sorry.

 

I wasn't trying to be mean or disrespect him. My reply does sound a bit harsh in retrospect haha. I didn't mean anything bad by it. I hope he knows that. Just trying to help and give my insight.

Benn.A  May 15, 2019
edited by Benn.A  May 15, 2019
 #6
avatar+102447 
+1

Yes I accept that   wink

Melody  May 15, 2019
 #2
avatar+14 
+2

= ma

F = 115 x 9.8m/s2

F = 1127N

This is due to Newton's first law stating that force equals mass times acceleration, the acceleration being the gravity of earth, 9.8m/s/s, and the mass being the acrobats ‘weight’ although it is actually his mass. The 130N applied to the horizontal wall is of no consequence to the force that is applied to the ground, while it would apply a torque force to his hand, this would be in Nm-1 and incalculable given the current information.

 May 15, 2019
 #7
avatar+18446 
+2

Hmmmmmm.....    I think I agree with Anthrax on this one.  The question did not ask for the NORMAL force on the hand....it asks for the magnitude of the  NET force on the hand.....  Ben, you even stated that the 130 N results in increased frictional FORCES on the hand.....   the RESULTANT force acting on the hand is not just the normal force of 1127 , but ALSO the horizontal force of 130 N and these two forces DO add as Anthrax calculated.....and they act in the direction of the hypotenuse (i.e. angled....not NORMAL).....   My one up vote goes with Anthrax.

  My 2 cents worth cheeky

 May 15, 2019

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