Physics Help Forum Hammer and nail torque

 Kinematics and Dynamics Kinematics and Dynamics Physics Help Forum

 Aug 10th 2017, 06:02 AM #1 Junior Member   Join Date: May 2017 Location: Sri Lanka Posts: 27 Hammer and nail torque Can you help me on solving this problem. θ=30 and when F=150N nail tries to remove. The force exerts by the nail is paralleled to the hammer. 1 Find the force exert by the single point of contact? 2.Find the force exert by the hammer to the nail? Attached Thumbnails
 Aug 10th 2017, 04:59 PM #2 Senior Member   Join Date: Apr 2015 Location: Somerset, England Posts: 995 There are four forces acting on the hammer. When the nail is about to move these forces maintain equilibrium and you can write the 3 equations of equilibrium involving these four forces and the geometry of the diagram. Have you tried this?
Aug 10th 2017, 09:15 PM   #3
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Join Date: May 2017
Location: Sri Lanka
Posts: 27
 Originally Posted by studiot There are four forces acting on the hammer. When the nail is about to move these forces maintain equilibrium and you can write the 3 equations of equilibrium involving these four forces and the geometry of the diagram. Have you tried this?
I tried to solve the problem.

I took the force which is exert by the nail on the hammer as Fn ,

The perpendicular distance to the nail force (Fn) from the single point of contact-5sin60

when we take the torque on the single point of contact,

150*30=5sin60*Fn
150*30=5*√3/2*Fn
Fn=1800/√3

Am I correct?

Aug 11th 2017, 04:17 AM   #4
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Join Date: Apr 2015
Location: Somerset, England
Posts: 995
 Originally Posted by osalselaka I tried to solve the problem. I took the force which is exert by the nail on the hammer as Fn , The perpendicular distance to the nail force (Fn) from the single point of contact-5sin60 when we take the torque on the single point of contact, 150*30=5sin60*Fn 150*30=5*√3/2*Fn Fn=1800/√3 Am I correct?
Are you?

You didn't comment on my suggestion.

Consider horizontal force equilibrium.

Does F = the horizontal component of Fn,

Which gives a different equation from your torque equation.

There must be one more force acting that you have not considered, can you tell me what it is?

 Aug 12th 2017, 04:29 AM #5 Junior Member   Join Date: May 2017 Location: Sri Lanka Posts: 27 Yes , According to the horizontal equilibrium Fnsin30=F. Therefore Fnsin30=150 Fn=150*2=300N Is the forth force you saying is gravity force of the hammer?
 Aug 13th 2017, 04:23 PM #6 Senior Member   Join Date: Apr 2015 Location: Somerset, England Posts: 995 Here is a force diagram using only three forces acting on the hammer. R is the (angled downwards) reaction of the nailhead on the claw. V is the vertical reaction at C, the point of contact of the hammerhead on the table. It is vertical because the table is shown horizontal. Simple horizontal equilibrium calculation yields R = 300kN But using the geometry of triangle ABC and calculating the moment equilibrium about C we find that R works out at 1039kN So something is wrong. Given that I have told you that you need a fourth force, can you see what this is? Attached Thumbnails
 Sep 12th 2017, 10:47 AM #7 Junior Member   Join Date: May 2017 Location: Sri Lanka Posts: 27 Reply Sorry for replying after long session. After your answer I tried very much for find the forth force. But I couldn't find it. What is that force. I have a big curiosity with it.
 Sep 12th 2017, 11:41 AM #8 Senior Member   Join Date: Apr 2015 Location: Somerset, England Posts: 995 It's good you maintained the interest. I see I made a silly mistake as well. I thought F was given in kN but it is just in N so my figures are 1000 times too large. Anyway the fourth fource. This acts at the point of contact between the hammer head and the table. That is the point of action of V on the hammer head. It is of course the horizontal force of friction. This shows that the claw hammer nail extractor will not work without friction. Without friction the hammer head would slide along the table. This friction force acts through the point about which I have taken moments so has zero moments about this point. So the moment equilibrium is correct. In order to find the friction required you add this into the equation for horizontal equilibrium So R sin(theta) = F + Friction (The frition stops the hammer head skidding to the left away from F and so acts in the same direction as F) 1039/2 = 150 + friction Friction = 520 - 150 = 370 N osalselaka likes this.
 Sep 13th 2017, 02:03 AM #9 Junior Member   Join Date: May 2017 Location: Sri Lanka Posts: 27 Thank you very much sir. You have given me a big point which I have mistaken for a whole month. Great help from you!!!!!!!!!! Thanks... a lot
 Sep 13th 2017, 05:32 AM #10 Senior Member   Join Date: Apr 2015 Location: Somerset, England Posts: 995 Happy to help. A further tip is this little theorem. If a (rigid) body is in equilibrium under the action of 3 forces only, then the lines of action of those 3 forces must meet at a point. This point may within or outside the body. Clearly, since they all pass through this point there are no moment equations so horizontal and vertical equilibrium alone can solve the problem. That is how I knew there had to be a fourth force, since the three force do not meet at a point.

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