Question : An engine mechanism is shown in Fig. 8.5. The crank CB = 100...

Given :     \omega_{ BC }=75 rad / s ; \alpha_{ BC }=1200 rad / s ^{2} , CB = 100 mm = 0.1 m; B A = 300 mm = 0.3 m

We know that velocity of B with respect to C or velocity of B,

v_{ BC }=v_{ B }=\omega_{ BC } \times C B=75 \times 0.1=7.5 m / s                                                                 …(Perpendicular to BC)

Since the angular acceleration of the crankshaft, \alpha_{ BC }=1200 rad / s ^{2} , therefore tangential component of the acceleration of B with respect to C,

Note: When the angular acceleration is not given, then there will be no tangential component of the acceleration.

1. Velocity of G and angular velocity of AB

First of all, draw the space diagram, to some suitable scale, as shown in Fig. (a). Now the velocity diagram, as shown in Fig.(b), is drawn as discussed below:

1. Draw vector cb perpendicular to CB, to some suitable scale, to represent the velocity of B with respect to C or velocity of B (i.e. v_{ BC } \text { or } v_{ B }), such that

2. From point b, draw vector ba perpendicular to B A to represent the velocity of A with respect to B i.e. v_{ AB } , and from point c, draw vector ca parallel to the path of motion of A (which is along AC) to represent the velocity of A i.e. v_{ A }.The vectors ba and ca intersect at a.

3. Since the point G lies on A B, therefore divide vector ab at g in the same ratio as G divides A B in the space diagram. In other words,

A B in the space diagram. In other words,

The vector cg represents the velocity of G.

By measurement, we find that velocity of G,

From velocity diagram, we find that velocity of A with respect to B,

We know that angular velocity of A B,

2. Acceleration of G and angular acceleration of AB

We know that radial component of the acceleration of B with respect to C,

and radial component of the acceleration of A with respect to B,

Now the acceleration diagram, as shown in Fig. (c), is drawn as discussed below:

1. Draw vector c^{\prime} b^{\prime \prime} parallel to CB, to some suitable scale, to represent the radial component of the acceleration of B with respect to C,i.e. a_{ BC }^{r}, such that

2. From point b^{\prime \prime}, draw vector b^{\prime \prime} b^{\prime} perpendicular to vector c^{\prime} b^{\prime \prime} or CB to represent the tangential component of the acceleration of B with respect to C i.e. a_{ BC }^{t}, such that

\text { vector } b^{\prime \prime} b^{\prime}=a_{ BC }^{t}=120 m / s ^{2} …(Given)

3. Join c^{\prime} b^{\prime}. The vector c^{\prime} b^{\prime} represents the total acceleration of B with respect to C i.e. a_{ BC }.

4. From point b^{\prime}, draw vector b^{\prime} x parallel to B A to represent radial component of the acceleration of A with respect to B i.e. a_{ AB }^{r} such that

5. From point x, draw vector xa’ perpendicular to vector b’x or B A to represent tangential component of the acceleration of A with respect to B i.e. a_{ AB }^{t}, whose magnitude is not yet known.

6. Now draw vector c^{\prime} a^{\prime} parallel to the path of motion of A (which is along AC) to represent the acceleration of A i.e. a_{ A }.The vectors x a^{\prime} and c^{\prime} a^{\prime} intersect at a^{\prime}. Join b^{\prime} a^{\prime}. The vector b^{\prime} a^{\prime} represents the acceleration of A with respect to B i.e. a_{ AB }.

7. In order to find the acceleratio of G, divide vector a^{\prime} b^{\prime} in g^{\prime} in the same ratio as G divides B A in Fig. (a). Join c^{\prime} g^{\prime}. The vector c^{\prime} g^{\prime} represents the acceleration of G.

By measurement, we find that acceleration of G,

From acceleration diagram, we find that tangential component of the acceleration of A with respect to B,

a_{ AB }^{t}=\text { vector } x a^{\prime}=546 m / s ^{2}              …(By measurement)

\therefore      Angular acceleration of A B,