Turning effects of forces

What is the centre of gravity of an object?

A. The geometrical centre of the object

B. The point about which the total torque is zero

C. The point at which the weight of the object may be considered to act

D. The point through which gravity acts

Answer: C

A. Incorrect: This is only true if the object is uniform in density and shape

B. Incorrect: This refers more accurately to the centre of mass in certain static conditions, or it might apply in specific contexts like rotational equilibrium

C. Correct: The centre of gravity is defined as the point at which the entire weight of an object can be considered to act

D. Incorrect: Gravity acts on every particle of the object, not just through one point. Saying it acts "through" a single point is misleading - what we consider is that the entire gravitational force is equivalent to a single force acting at the centre of gravity

This diagram shows a box of length 25 cm acted on by four forces:

a. Which two forces constitute a couple acting on the box? Explain your choice.
b. Calculate the torque of the couple.

a.  forces B and D; equal magnitudes, opposite directions, not in same straight line

b. `"torque of the couple:" " " tau = F xx d`

`tau = 20 " " N xx 0.25 " " m = 5 " " N.m`

A force F is applied at a distance d from the hinge H and an angle x to the door.

What is the moment of the force F about the point H?

The force `F` is at an angle `x` to the direction of the door (horizontal)

So, the perpendicular component of the force is `F_⟂ = F xx sin x`

`tau = F_⟂ xx d = F xx d xx sin x`

Some weights are hung from a light rod AB as shown in figure. The rod is pivoted. Calculate the magnitude of the force F required to balance the rod horizontally.

Sum of clockwise moments = `(0.25 × 1.2) + 0.35F`
Sum of anticlockwise moments = `0.40 × 2.5`
By the principle of moments
`(0.25 × 1.2) + 0.35F = 0.40 × 2.5`
`0.35F = 1.0 - 0.3`
`F = 2.0 " " N`

The beam shown in figure is in equilibrium. Determine the force X.

Determine the clockwise moments:

`"sum of moments of clockwise forces" = (10 xx 1.0) + (20 xx0.5) = 10 + 10 = 20 " " N.m`

`"moment of anticlockwise force" = X × 0.8`

Since we know that the beam must be balanced, we can write:
sum of clockwise moments = sum of anticlockwise moments

`20 = X xx 0.8xx X = 20 / 0.8 = 25 " " N`

So a force of 25 N at a distance of 0.8 m from the pivot will keep the beam still and prevent it from rotating (keep it balanced).

A uniform rod XY of length 0.80 m is at an angle of 40° to a horizontal surface, as shown in figure

Four forces A, B, C and W act on the rod to maintain equilibrium. Force C acts at right angles to the rod at end X and is 2.0 N. A vertical force B and a horizontal force A act on the rod at end Y. W is the weight of the rod.
a. Show that the weight of the rod is 5.2 N by taking moments about end Y.
b. State the name of the force B and A.

a. `"lever arm for weight"=0.40 xx cos(40^o) = 0.40 xx 0.766 = 0.3064 " " m`

`"moment of weight"=0.3064xxW`

`"moment of force C"=2xx0.8=1.6 " " N.m`

So, `0.3064xxW=1.6`

`W="5.2 "N"`

b. 

Force B: normal force

Force A: friction

A 30 cm beam is pivoted at its centre. Equal and opposite forces of 20 N pull the ends of the beam. What magnitude is the torque of the couple?

`"torque of the couple" = "force" xx "separation of forces" = 20 xx 0.3 " " ("distance in meter") = 6 " " N.m`

The torque produced by a person using a screwdriver is 0.18 N.m. This torque is applied to the handle of diameter 4.0 cm. Calculate the force applied to the handle.

`"force" = "torque"/"separation of forces" = 0.18 / (0.04) = 4.5 " " N`

In figure, a light rod AB of length 45 cm is held at A so that the rod makes an angle of 65° to the vertical. A vertical force of 15 N acts on the rod at B. Calculate the moment of the force about the end A

`"moment of force" = "force" xx "perpendicular distance from pivot"`

`15xx0.45sin65="6.1 N.m"`

A rigid uniform beam of weight W is connected to a fixed support by a hinge, as shown in figure

a.  Show that the vertical component of the tension in the string is 2.4 N

b.  By taking moments about the hinge, determine the weight W of the beam.

c.  Calculate the horizontal component of the force exerted on the beam by the hinge.

a.

`"vertical component of tension" " " T_y=4.8xxsin30^o="2.4 N"`

b.

`"clockwise moments"=0.3xx(0.5+0.1+0.2)+Wxx(0.5+0.1)+T_yxx(0.4+0.2+0.1+0.5)`

`"anticlockwise moment"=8.2xx0.5`

`"clockwise moments" " " = " " "vertical moment"`

`0.3xx(0.8)+Wxx(0.6)+T_yxx(1.2)=8.2xx0.5`

`W="1.6 N"`

c. 

`"horizontal force"=4.8xxcos30^o="4.2 N"`