Question 1 of 7: Three short mechanism questions

Question 1: Three short mechanism questions (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Three independent parts on planar-mechanism fundamentals: (1.1) synthesise a 5-link chain with one degree of freedom that contains exactly one higher pair; (1.2) a slider-crank of RRRP type (revolute–revolute–revolute–prismatic) with slider/cylinder friction $\mu_k=0.25$; (1.3) the mobility of a hydraulically-actuated air-stair deployment linkage.

Find. (1.1) two distinct sketches; (1.2) the meaning of $\beta$ and a force-transmission verdict over $-14^\circ<\beta<14^\circ$; (1.3) the number of degrees of freedom by Gruebler / Kutzbach.

1.1 — Two one-DOF five-bar mechanisms with one higher pair

Kutzbach's criterion with higher pairs is $M=3(n-1)-2J_1-J_2$, where $n$ is the number of links (including ground), $J_1$ the number of full (low-pair) joints and $J_2$ the number of half (higher-pair) joints. A bare five-bar linkage ($n=5$, five revolutes) has $M=3(4)-2(5)=2$. Replacing the "sixth" constraint by a single higher pair removes one freedom:

$$M=3(5-1)-2(5)-1(1)=12-10-1=\boxed{1}$$

so any five-link loop closed by five revolutes plus one higher pair is a one-DOF mechanism. Two valid, physically distinct realisations:

Both have $n=5$, $J_1=5$, $J_2=1\Rightarrow M=1$.

1.2 — RRRP transmission angle and self-locking

Here $\beta$ is the angle the coupler (link 3) makes with the normal to the slider path — equivalently, $90^\circ$ minus the transmission angle at the slider. Because link 3 is a two-force member, the force it delivers to the slider acts along link 3, i.e. inclined $\beta$ from the guide normal. Resolving that coupler force $F$ at the slider:

• driving component along the guide: $F\sin\beta$;
• component pressed into the guide (produces the normal reaction): $N=F\cos\beta$.

Motion of the slider requires the driving component to exceed the friction force $\mu_k N$:

$$F\sin\beta > \mu_k\,F\cos\beta \;\;\Longrightarrow\;\; \tan\beta > \mu_k .$$

The friction angle is

$$\varphi=\arctan\mu_k=\arctan(0.25)=\boxed{14.0^\circ}.$$

Therefore, throughout the range $-14^\circ<\beta<14^\circ$ we have $|\tan\beta|<0.25=\mu_k$, so the driving component $F\sin\beta$ is smaller than the maximum available friction $\mu_k F\cos\beta$. The line of action of the coupler force falls inside the friction cone: the mechanism self-locks.

Verdict: for $-14^\circ<\beta<14^\circ$ no power (neither force nor motion) can be transmitted from link 2 to link 4 — any input torque merely jams the slider against its cylinder. Transmission becomes possible only once $|\beta|$ exceeds the $14^\circ$ friction angle. (The examiners set the stated bound at $\pm14^\circ$ precisely because $\arctan 0.25=14.04^\circ$.)

1.3 — Mobility of the air-stair mechanism

Reading the figure: the rod protrudes from the top of the actuator and is pinned to the upper ground mount, the cylinder body carries the mid-span pin that takes the upper support link back to ground, and the cylinder's lower end pins to the tread. Enumerating the bodies and joints:

So $n=6$ links and $J_1=7$ lower pairs (six revolutes + one prismatic), with no higher pairs, $J_2=0$:

$$M=3(n-1)-2J_1-J_2=3(6-1)-2(7)-0=15-14=\boxed{1}.$$

The linkage has one degree of freedom, consistent with its single hydraulic-actuator input driving the entire stair between the stowed and deployed positions.