Question

Solve question no. 22

Answer 1

Answer:

Let F_(1)=10 N , r_(1)=3m, F_(2)=8 N and r_(2)=4 m

We know that Moment = $F.r.sin\theta$

Case 1

Moment =$F_{1}.r_{1}.sin\: \theta$

$=30.sin \theta$

Case 2

Moment = $F_{2}.r_{2}.sin\: \theta$

$=32.sin \theta$

Since, 32.sin theta>30 sin theta,

8 N force with arm of 4 m produces greater moment

Answer 2

Answer:

Moment of Force is also called as Torque . It's an axial vector denoted by the Greek Letter "Tau".

The basic relationship of torque with force is that :

• Torque is known as the moment of force , which means that it is defined as a vector product of position vector with Force vector

Mathematically , we can say that :

$\boxed{ \blue{ \bold{ \huge{ \tau = | \vec f| \times | \vec r| \times \sin( \theta) }}}}$

θ refers to the angle between the Force vector and the position vector from the axis.

Calculation:

In first case , the force is 10 N and distance of torque application from axis is 3 m

$\boxed{\tau1 = 10 \times 3 \times \sin(\theta)= 30\sin(\theta) \: Nm }$

In the 2nd case , the force is 8N , and distance of application of torque from axis is 4

$\boxed{ \tau2 = 8 \times 4\times \sin(\theta) = 32 \sin(\theta) \: Nm }$

So moment of force is greater on 2nd case ✓