Lumberjacks use cranes and giant tongs to hoist their goods into trucks for transport. Fortunately, smaller versions of these devices are available for weekend warriors who want to play with their chainsaws.

Let us model the illustrated tongs as a planar mechanism that carries a log of weight \(N\).

Given the following dimensions:
- 35 mm
- 10 mm
- 40 mm
- 230 mm
- 85 mm
- 45 mm
- 10 mm
- 35 mm
- 345 mm

Determine:
1. The force in \(N\) and moment in \(Nm\) that our worker exerts on the tongs.
2. The pinching force magnitude in \(N\) that the tongs exert on the log; i.e., determine the horizontal force that the tong's teeth exert on the log. Assume the point is centered between the tong's teeth.

Matlab input: a

Lumber jacks use cranes and giant tongs to hoist their goods into trucks for transport. Fortunately, smaller versions of these devises are available for weekend warriors who want to play with their chain saws. Let us model the illustrated tongs as a planar mechanism that carries a log of weight 210 N. Given the following dimensions: 35 mm 10 mm 40 mm 230 mm 85 mm 45 mm 10 mm 35 mm 345 mm determine the force in N and moment in Nm that our worker exerts on the tongs. Also determine the pinching force magnitude in N that the tongs exert on the log; i.e. determine the horizontal force that the tong's teeth exert on the log. Assume the point E is centered between the tong's teeth.

The diagram for this question is shown on the first uploaded image

Answer:

The force P is [tex]P= 210 N[/tex]

and the moment M is [tex]M = -48.3N \cdot m[/tex]

The horizontal force that the tong teeth exerts is [tex]F_T =89.67N[/tex]

Explanation:

First let denote the dimension to corresponding to the diagram

[tex]a=35mm , b= 10mm, c= 40mm, d= 230mm, e= 85mm,f= 45mm,\\g= 10mm,h=35mm,i=345mm.[/tex]

Next looking at the diagram let us consider the vertical direction

At equilibrium

[tex]\sum F_{vertical} =0[/tex]

This mean that

[tex]P+ W = 0[/tex]

Since they are acting in opposite direction the equation becomes

[tex]P - W = 0[/tex]

=> [tex]P= W[/tex]

=> [tex]P= 210 N[/tex]

At Equilibrium Moment about F gives

[tex]\sum M_f = 0[/tex]

=> [tex]F_T * (e +f + g+ h+ i) - F_T * (e+ f+g+ h+i) - W *d -M =0[/tex]

=> [tex]M = -W *d[/tex]

=> [tex]M = -210 * 0.230[/tex]

=> [tex]M = -48.3N \cdot m[/tex]

Here [tex]F_T[/tex] is the horizontal force that the tong teeth exerts

Now let consider the part BAF of the system as shown on the second uploaded image

Now the angle [tex]\theta[/tex] is mathematically given as

[tex]tan \theta = \frac{g+h}{a}[/tex]

=> [tex]\theta = arctan \frac{g+h}{a}[/tex]

[tex]= arctan (\frac{10+35}{35} )[/tex]

[tex]=52.125^o[/tex]

Now at equilibrium the moment about A is

[tex]\sum M_A = 0[/tex]

=> [tex]P * (c+d) +M + F_{BC} cos \theta *f+F_{BC}sin\theta *(a+b) =0[/tex]

[tex]210 * (0.040 + 0.230)-48.3+F_{BC} cos (52.125^o)*0.045+ F_{BC}sin(52.125^o)[/tex][tex]* (0.035 +0.010) =0[/tex]

=> [tex]10.29 +F_{BC} (0.02763+0.03552) =0[/tex]

=> [tex]F_{BC} =\frac{10.29}{0.06315}[/tex]

=> [tex]F_{BC} = - 162.925 N[/tex]

Looking at the forces acting on the teeth as shown on the third uploaded image

At Equilibrium the moment about D is

[tex]\sum M_D = 0[/tex]

=> [tex]\frac{W}{2} *d - F_T *(i+h) -F_{BC} cos \theta *h -F_{BC} sin \theta * (b+c) =0[/tex]

=> [tex]\frac{210}{2} * 0.230 -F_T (0.345 +0.035) - (-162.925)cos(52.125^o) *0.035\\-(-162.925)sin(52.125^o)(0.010 +0.040) =0[/tex]

=> [tex]34.081 = F_T(0.345 +0.035)[/tex]

=> [tex]F_T =89.67N[/tex]