PRINCIPLES OF DAIRY MACHINE DESIGN

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Module 1. Statics and Dynamics

Lesson 3
SOLVING NUMERICALS

3.1 Problem

(OR)

Adding equations (1) and (2), we get

(P+Q) ² = P² + Q² + 2PQ = 10+6 = 16

P+Q = 4 → (3)

Similarly,

(P-Q) ² = P² + Q² - 2PQ = 10 - 6 = 4

P-Q = 2 → (4)

(OR) Solving the equations 3 & 4

P = 3 N

Q = 1N

3.2 Problem

Two forces act at an angle of 120⁰. The bigger force is of 40 N and the resultant is perpendicular to the smaller one. Find the smaller force.

Fig. 3.3

Sol.

Bigger force P = 40 N

Angle between the resultant and the Smaller force = 90⁰

Angle between the two forces θ = 120⁰

Angle between the bigger force and the resultant, Α=120-90=30⁰

Let Q = Smaller force

3.3 Problem

A body of mass100 N is suspended by two strings of 4 m & 3m lengths attached at the same horizontal level of 5 m a part. Find the tensions in strings.

Fig. 3.4

3.4 Problem

A component of weight of 50 N is hauled along a rough horizontal plane, by a pull of 18 N acting at an angle of 14⁰ with the horizontal. Find co-efficient of friction.

Fig. 3.5

Sol.

Weight W = 50 N

Inclination of force

θ = 14⁰

Force P = 18 N

Let R = Normal reaction

µ = co-efficient of friction

F = Force of friction

Resolving the forces at right angles to the plane

R= 50-18 Sin 14⁰ = 45.65 N → (1)

Now resolving the force along the plane

F = 18 Cos 14⁰ = 17. 46 N → (2)

We, know that

F = µ R

3.5 Problem

A force of 40 N pulls a component of weight 60 N up an inclined of the plane, to the horizontal, is 30⁰, calculate the co-efficient of friction.

Fig. 3.6

Sol.

Given weight, W = 60 N

Force P = 40 N

Inclination θ = 30⁰

Let R = Normal reaction

µ = co-efficient of friction

F = Force of friction

Resolving forces along the inclined plane

F = 40-60 Sin 30⁰ = 10 N → (1)

Resolving force at right angles to the plane

R = 60 Cos 30⁰ = 51.26 N → (2)