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.
Horizontal component of the tension
= 3000 cos30° =2598N
Friction = 500N
Resultant horizontal force
= 2598N – 500N =2098N
The 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.
Horizontal component of the forces
= 25cos45° + 25cos45° = 35.36N
Vertical component of the forces
= 25sin45° – 25sin45° = 0N
The acceleration of the object can be determined by the equation
F = ma
(35.36) = (2)a
a = 17.68 ms-2