Test: Moment Of A Force - 2 - Civil Engineering (CE) MCQ

The curled hand represents various thing. The direction of the moment axis is given by the thumb. The direction of the force is given by the fingers. As we place the fingers on the force and curl towards the rotational direction of the body about the axis.

As we know that the moment is the cross product of the distance and the force we will try to apply the same here. But here we have number of forces components acting on the T point. So by adding all the moments caused by all the forces will give us the value as 0Nm.

As we know that the moment is the cross product of the distance and the force we will try to apply the same here. We see that the perpendicular distance is 3sin60. Thus we get the distance. And hence multiply it with the force, moment = 112.5Nm. Because the force component perpendicular to the distance needs to be taken.

When we consider about the dimensions we need to be careful.The moment is the vector quantity. Thus the value of the total moment caused by various forces acting on the body is the vector sum of all the vectors. Also the moments are not perpendicular to each other, unless it is specified.

The moment axis is always perpendicular to the planes of the force and the distance of the axis and the point of action of the force on the body. This means that the moment is the cross product of the force and the distance between the axis and the point of action of the force.

As we know that the moment is the cross product of the force and the distance between the point of contact of the force and the point about which moment needs to be calculated. Here distance r1 =5j m and r2 = 4i + 5j – 2k m. Thus doing the cross product will give the answer.

The moment of the force about the axis of rotation by the application of the force on the body is given by the cross product of both. If the force not perpendicular to the axis, and making angle θ then cosine form of angle is used. As usually used in the cross product.

The moment of the force about the axis of rotation by the application of the force on the body is given by the cross product of both. If the force not perpendicular to the axis, and making angle θ then cosine form of angle is used. Thus, FLsinθ represents the magnitude of the moment.

The basic way of doing so is to use right hand rule and not the left hand rule. The direction of the moment axis is given by the thumb. The direction of the force is given by the fingers. As we place the fingers on the force and curl towards the rotational direction of the body about the axis.

If you will apply he right hand rule on the system given then the right answer is the positive z-axis. Which is because the force is lying on the x-axis and is heading towards the positive infinity of the x-axis. And the points are in the positive y-axis. Apply the rule, and get the direction.

If a force applied at any point in its line of action and is still creating the same moment about any fixed point say P, then the force is said to be sliding vector. This is because the moment of the force which is acting on its line of axis at the point P is same throughout. Whatever be the direction of the distance.

 If a force applied at any point in its line of action and is still creating the same moment about any fixed point say P, then the force is said to be sliding vector. This is because the moment of the force which is acting on its line of axis at the point P is same throughout. This is known as the principle transmissibility of the force.

Yes, we can prepare the moment in the Cartesian form. As the moment in the 3D is the vector. Which can be easily made in the form of Cartesian coordinates. Also it can be seen that the moment is the cross product of the force and the distance, hence the moment is in vector form.

The given equation represents the total moment of the forces which are acting on the body. That is the summation of all the rxF. Where the r is the distance of the axis from the point of action of the force on the body. And thus this is the total summation of the moments of all the forces acting on the body.

The force developed is 2KN, and the roots are having the coordinate (0, 0, 0). Coordinates of the top of the tree is (0, 0, 12). The tractor’s coordinates are (4, 12, 0). Thus applying the cross product on the force and the distance of the tractor from the roots we get the answer as -16.5i+5.51j KNm.

If F = F1+F2, then moment of this force F about a point at a distance r is M=rxF1 + rxF2.
As M = rx(F) = rx(F1 + F2) = rxF1 + rxF2. This is known as the principle of moments. As the force is a vector quantity thus this is the distributive property which we apply to get the answer.

As we know that the moment is the cross product of the force and the distance between the point of contact of the force and the point about which moment needs to be calculated. Thus forming the distance vector and then crossing it with the force will give us the answer. Remember force also needs to be in the vector form for doing the cross product.

As we know that the moment is the cross product of the force and the distance between the point of contact of the force and the point about which moment needs to be calculated. Thus forming the distance vector and then crossing it with the force will give us the answer. Remember force also needs to be in the vector form for doing the cross product.

As we know that the moment is the cross product of the force and the distance between the point of contact of the force and the point about which moment needs to be calculated. Thus forming the distance vector and then crossing it with the force will give us the answer. Remember force also needs to be in the vector form for doing the cross product.

As we know that the moment is the cross product of the force and the distance between the point of contact of the force and the point about which moment needs to be calculated. Here distance r1 =5j m and r2 = 4i + 5j – 2k m. Thus doing the cross product will give the answer.