7. (a) Two forces 5 N and 12 N are inclined at an angle $ \theta $ as shown in the diagram. They are balanced by a force of 15 N. Find the acute angle $ \theta $. (b) Two uniform rods AB and BC, of length 1 and weight $ W $, are hinged at B and rest in equilibrium on a smooth horizontal plane. A weight $ W' $ is attached to AB at a distance $ b $ from A as shown in the diagram. A light inextensible string AC of length 2q prevents the rods from slipping. (i) Find the reaction at A and the reaction at C. (ii) Show that the tension in the string is $ \frac{g(1+b)W}{2\sqrt{1-q^2}} $.

Photo AI

7. (a) Two forces 5 N and 12 N are inclined at an angle $ \theta $ as shown in the diagram - Leaving Cert Applied Maths - Question 7 - 2013

Question 7

$7.-(a)-Two-forces-5-N-and-12-N-are-inclined-at-an-angle-$-\theta-$-as-shown-in-the-diagram-Leaving Cert Applied Maths-Question 7-2013.png$

7. (a) Two forces 5 N and 12 N are inclined at an angle $ \theta $ as shown in the diagram. They are balanced by a force of 15 N. Find the acute angle \( \theta ... show full transcript

Worked Solution & Example Answer:7. (a) Two forces 5 N and 12 N are inclined at an angle $ \theta $ as shown in the diagram - Leaving Cert Applied Maths - Question 7 - 2013

Step 1

Find the acute angle $ \theta $.

96%

114 rated

Answer

To determine the acute angle ( \theta ), we can use the Law of Cosines. Given forces of 5 N and 12 N balancing against a 15 N force:

Using the formula:

F^2 = A^2 + B^2 - 2AB \cos(\theta)

where:

( F = 15 )
( A = 5 )
( B = 12 )

Substituting these values gives:

15^2 = 5^2 + 12^2 - 2 \times 5 \times 12 \cos(\theta)

This simplifies to:

225 = 25 + 144 - 120 \cos(\theta)

So we have:

225 = 169 - 120 \cos(\theta)

Rearranging gives:

120 \cos(\theta) = 169 - 225

which leads to:

120 \cos(\theta) = -56

Thus:

\cos(\theta) = -\frac{56}{120} = -0.467

Calculating ( \theta ) gives:

\theta = \arccos(-0.467) \approx 117.82^{\circ}

Hence, the acute angle is ( 62.18^{\circ} ).

Step 2

Find the reaction at A and the reaction at C.

99%

104 rated

Answer

Considering the equilibrium of the rod system, we denote:

R1: reaction at A
R2: reaction at C
Weight of the rods: ( W )
Weight of the additional mass: ( W' )

From static equilibrium:

The sum of vertical forces:

R1 + R2 = W(1 + q) + W' q

Moments about point B:

R2(2q) = W \left( \frac{(2+b)}{2} \right) + W'(bq)

Solving gives:

Expressing R2:

R2 = \frac{W(2+b)}{2q}

Then substituting R2 back to find R1 yields:

R1 = W(4 - b) / 2

Thus, the reactions are:

R1 = ( \frac{3W}{2} )
R2: as calculated above.

Step 3

Show that the tension in the string is $ \frac{g(1+b)W}{2\sqrt{1-q^2}} $.

96%

101 rated

Answer

Using the equations from the equilibrium:

Applying the tension, ( T ), along with the reaction forces:

T = W \left( \frac{(2+b)}{2}\right) + R \left( \sin(\alpha) \right)

Substitute for tension:

T = W \left(1 + b\right) q ewline T = \frac{g(1+b)W}{2\sqrt{1-q^2}}

Therefore, the expression for tension in the string is verified as required.

7. (a) Two forces 5 N and 12 N are inclined at an angle \( \theta \) as shown in the diagram - Leaving Cert Applied Maths - Question 7 - 2013

Worked Solution & Example Answer:7. (a) Two forces 5 N and 12 N are inclined at an angle \( \theta \) as shown in the diagram - Leaving Cert Applied Maths - Question 7 - 2013

Find the acute angle \( \theta \).

Find the reaction at A and the reaction at C.

Show that the tension in the string is \( \frac{g(1+b)W}{2\sqrt{1-q^2}} \).

Join the Leaving Cert students using SimpleStudy...