Forces in Equilibrium

Answer:

            \begin{gathered}
  {\text{Magnitude of the resultant force,}} \hfill \\
  |F| = \sqrt {10^2  + 12^2 }  = \sqrt {244}  = 15.62N \hfill \\ 
\end{gathered} 
            

Diagram above shows that four forces of magnitude 2N, 4N, 5N and 8N are acting on point O. All the forces are perpendicular to each others. What is the magnitude of the resulatant force that acts on point O?

Answer:

The resultant force of the horizntal component = 5 - 2 = 3N to the right
The resultant force of the vertical component = 8 - 4 = 4N acting downward.

Therefore, the magtitude of these 2 force components,

            |F| = \sqrt {3^2  + 4^2 }  = \sqrt {25}  = 5
            

Diagram above shows a lorry pulling a log with an iron cable. If the tension of the cable is 3000N and the friction between the log and the ground is 500N, find the horizontal force that acting on the log.

Answer:

Horizontal component of the tension = 3000 cos30o =2598N
Friction = 500N

Resultant horizontal force = 2598N - 500N =2098N

Diagram above shows two forces of magnitude 25N are acting on an object of mass 2kg. Find the acceleration of object P, in ms-2.

Answer:

Horizontal component of the forces = 25cos45o + 25cos45o = 35.36N

Vertical component of the forces = 25sin45o - 25sin45o = 0N

The acceleration of the object can be determined by the equation

F = ma
(35.36) = (2)a
a = 17.68 ms-2

A block of mass 2 kg is pulling along a plane by a 20N force as shown in diagram above. Given that the fiction between block and the plane is 2N, find the magnitude of the resultant force parallel to the plane.

Answer:

First of all, let's examine all the forces or component of forces acting along the plane.

The force pulling the block, F = 20N
The frictional force Ffric = 2N
The weight component along the plane = 20sin30o = 10N

The resultant force along the plane = 20 - 2 - 10 = 8N

Diagram above shows a load of mass 500g is hung on a string C, which is tied to 2 other strings A and B. Find the tension of string A.

Answer:

Tension of string C, TC = weight of the load = 5N
All forces in the system are in equilibrium, hence

Vertical component of tension A (TA) = TC
TAcos60o = TC
TA = TC/cos60o
TA = 5/cos60o = 10N