Test: Frictional Forces On Flat Belts - Civil Engineering (CE) MCQ

The friction is the phenomena that defines that there is a resistance which is present there between the two surfaces. The two surfaces are in contact and the friction applies at that surface only, resisting the motion of the surface.

The friction is the phenomena that defines that there is a resistance which is present there between the two surfaces. This friction is applied tangentially to the surfaces in contact. Thus the main thing is that the forces on both of the surfaces act tangential to each other.

The sine and the cosine components of the given vectors considering the angle B as the only angle of consideration comes 1.5m and 2.59m.

For solving of the unknown tension in the belts, T₂ = T₁e^µB equation is used. In this the R.H.S tension is the maximum tension of the two tensions. While the other one is the smaller one. And the µ is coefficient of friction between the belt and the surface. And the B is the angle of belt to the surface in radians.

The dry friction is acted upon the surfaces. Whatever may be the surfaces. And they are tangential to each other. As we know the friction is the phenomena that defines that there is a resistance which is present there between the two surfaces. The dry friction is also termed as the Coulomb friction as it was given by C.A. Coulomb.

As the loads are being acting in the downward direction. Thus to make the forces balance, the normal forces act in the vertically upward direction. As we know that when there is no lubricating fluid present between the surfaces in contact, the dry friction occurs. This friction magnitude is taken out from these normal forces

For the equilibrium in the three dimensional system of axis we have all the conditions true as, ∑Fx=0, ∑Fy=0 and ∑Fz=0. Also we have the summation of the forces equal to zero. Which is not a non-zero value.

We first make the free body diagram and then we make the equilibrium equations to satisfy the given conditions. This helps us to solve the question easily. As this reduces the part of imagination and increases accuracy too.

The net forces acting on the body is shown by the help of the resultant forces. This is because the resultant forces have the sum of all the forces which are acting on the direction which is same. Thus the resultant forces are used to show the net forces acting in the body.

The net forces acting on the body is shown by the help of the resultant forces. There are two types, first the frictional and the second is the normal. This is because the resultant forces have the sum of all the forces which are acting on the direction which is same.

The component of the force in the v axis, it is equal to 100cos(15˚). This is the application of the triangle over the figure. Try to resolve the components of the given force. It will be easy. This is one of the simplest example of solving for the forces acting over the belts.

If a vector is multiplied by a scalar then its magnitude is increased by amount of that scalar’s magnitude. When multiplied by a negative scalar it going to change the directional sense of the vector. Vector math is unchanged throughout.

All the vectors quantities obey parallelogram law of addition. Two vectors A and B (can be called as component vectors) are added to form a resultant vector. R = A+B.

For solving of the unknown tension in the belts, T2 = T1eµB equation is used. In this the R.H.S tension is the maximum tension of the two tensions. While the other one is the smaller one. And the µ is coefficient of friction between the belt and the surface.

After getting the cosine components of the given vectors we obtain the total length of the x-axis components to be 3cos60 + 2cos30 = 3.23.