chapter 14 hw solution - university of new mexico

ME 314 Chapter 14 HW April 16, 2010

Chapter 14 HW Solution

Problem 14.2: The force on the slider is P = 900 N, and we want to find the moment M12 on link 2 to maintainequilibrium.

The free-body diagrams of bodies 2 and 3 are shown below.

The law of sines reveals that angle φ = 11.95◦ (we found this in class). The direction I chose for the forces (andmoment) is arbitrary, but from now on you must be consistent. Note that I chose M12 to be CCW (positive),which is opposite from text Figure P14.2.

In drawing the free-body diagram I’ve also used the fact that link 3 is a “3-force” member, so the lines of actions ofall forces must intersect (at point B). Hence we know the direction of F23.

Summing forces on link 3:

F23 (cosφi− sinφj)︸︷︷︸F23

+F13 j− P i = 0 (1)

From the i components we get

F23 =P

cosφ= 920 N, (2)

so that

F32 = −F23 = −920 (cosφi− sinφj) (3)

Next summing moments on link 2 (around point O2) gives

rAO2 × F32 + M12 = 0. (4)

From (4) we get the final result,

M12 = −rAO2× F32 = −61.5k N-m (5)

Since M12 is negative, it is in the CW direction, and opposite the direction indicated in the FBD above.


Figure 1: Quick-returnmechanism.

Problem 14.7: This is a “quick-return” mechanism. There is no friction, sothe constraint force between links 1 & 6 is normal to the line of contact, andthe constraint force between links 2 & 4 is likewise normal to the line of contact(link 4). Link 5 is a “2-force” member.

The analysis proceeds with the free-body diagram of link 6, then to 2-forcemember link 5, then to link 4, and finally to link 2 (body 3 simply transmitsthe static force between 2 and 4).

However, before one can perform this analysis, the various angles of the linksand the length O2Amust be determined using trigonometry and/or loop closureequations. The angle of link 2 is given as 30◦, but the angles of links 4 and 5must be found. Neglecting the details (I can show you if you wish), my resultswere:

φ4 = 16.6274◦ (angle of link 4 from vertical)

φ5 = 4.7972◦ (angle of link 5 from horizontal)

rO4A = 7.5664 in

• Links 6 & 5: The free-body diagrams for links 6 and 5 are shown below. Note that since link 5 is a 2-forcemember, the directions of F65 and F45 are known; also F56.

φ5 = 4.797◦

Using φ5, the force balance yields

P i + F16 j + F56(−0.9965i + 0.0836j) = 0 (6)

from which I found that

F16 = −8.3923 lb (7)

F56 = 100.3515 lb (8)

The negative value for F16 indicates that this force is really in the opposite direction from my drawing (i.e. it isdownwards). This is obvious if you examine Figure 1.

Also, for link 5, F65 = −F56, and F45 = −F65.


• Link 4: The free-body diagram for link 4 is shown below. We know the direction of both F54 and F24. Slider 3does not really enter into this static analysis—I’ll not show it either here or in Problem 14.15.

φ4

=16

.63◦

Force F54 is 100.3515 lb from before, and acts at the angle of φ5. Since there is no friction, force F24 is normalto link 4 at A.

There are two ways to proceed with the analysis of link 4:

– Consider link 4 as a 3-force member, so the line of action of all three forces interect at a common point

– Perform a moment balance around point O4

The first method is awkward here because the two lines of action of F54 and F24 are nearly parallel, hence theintersection point is quite distant. The second method is much more feasible. This moment balance is

MO4= rBO4

× F54 + rAO4× F24 = 0 (9)

which yields

−1571k + 7.5664F24k = 0 =⇒ F24 = 207.63 lb (10)

• Link 2: The final free-body diagram is of link 2, shown below. Both forces are completely known.

A

Summing moments around point O2 we obtain

MO2 = rAO2 × F42 + M12 = 0 (11)

which yields

−377.3155k +M12k = 0 (12)


so

M12 = 377k in-lb (13)

which is CCW.

Problem 14.15: This is the same mechanism as before, but now with friction acting at two places: between links 1 & 6(µc = 0.20), and between links 3 & 4 (µc = 0.10).

The ONLY difference in this analysis is that the contact forces between these bodies will no longer be normal, butwill be inclined at friction angle φ = tan−1 µc from normality.

The inclination of these interaction forces is due to the addition of a tangential friction force, which always opposesthe impending relative motion caused by moment M12 as it “drives” against force P .

• Moment M12 causes block 6 to slide to the LEFT relative to link 1 =⇒ friction force F t16 acts to the RIGHT

• Not-so-clearly, force P causes slider 3 (or 2) to move UP relative to link 4 =⇒ friction force F t42 acts DOWN

The friction angles are:

φ16 = tan−1 0.2 = 11.31◦

φ24 = tan−1 0.1 = 5.71◦

Procedure: In drawing the new FBDs, I will draw the normal component of the contact force in the direction itactually acts (known from Problem 7). By adding the tangential frictional forces in the directions indicated above,the resulting contact force will be thereby “inclined” by the friction angle from the normal direction.

Updated Free Body Diagrams. I’ve shown the updated free body diagrams below, with the forces “inclined” dueto friction. The inclinations (11◦ and 6◦) are fairly small.

• Links 6 & 5: The updated free-body diagram is shown below. Contact force F16 in the direction it “really”is—DOWN. Since the “impending” motion of the block due to moment M21 is to the left, the friction forcecomponent on the block will have a component to the right. The resultant contact force is thereby inclined tothe right by φ16 as shown (normal and tangential components are shown in blue).

F16φ16

The forces in the FBD are:

P = 100i (14)

F16 = F16 (sinφ16i− cosφ16j) (15)

F56 = F56 (− cosφ5i + sinφ5j) (16)

I got i and j equations of

100 + 0.1961F16 − 0.9965F56 = 0 (17)

−0.9806F16 + 0.0836F56 = 0 (18)

I solved these with MATLAB to get

F16 = 8.70 =⇒ F16 = 14.36i− 71.78j lb (19)

F56 = 102.06 =⇒ F56 = 101.71i + 8.54j lb (20)

With friction, both contact forces are slightly larger than before.


• Link 5: The simple free-body diagram for link 5 is shown below; forces equal and opposite and known.

• Link 4: The revised FBDs for links 2 and 4 are shown below. Since body 3 (or 2) moves UP relative to link 4,the tangential friction force component of F42 at point A will be DOWN. The reaction force F24 is easier todraw—the tangential friction component goes UP (thin blue lines for components). So F24 is inclined as shown;force F42 is equal and opposite.

F54

F14

F42

F24

φ24

Force F54 is known from F56, and is

F54 = 101.7060i + 8.5354j (21)

Since link 4 is 16.6724◦ from vertical, and F24 is φ24 CW from the normal to link 4 (see FBD), the angle of F24

is 22.38◦ above the horizontal. So the expression for force F24 is

F24 = F24(cos 22.38◦i− sin 22.38◦j) = F24(0.9247i− 0.3807j) (22)

Compute moments around O4 as before, so

MO4= rBO4

× F54 + rAO4× F24 = 0 (23)

−1598.3k + 7.528F24k = 0 (24)

so we find that

F24 = 212.3 lb =⇒ F24 = −196.3i + 80.8j lb (25)

• Link2: Referring to the FBD for link 2, we simply take moments around O2, which yields

M12 − rAO2 × F42 = 0 (26)

So the final result is

M12 = 420.4k in-lb (27)

This is significantly larger than the previous value of 377 in-lb, illustrating the contribution of friction.

chapter 14 hw solution - university of new mexico

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