Question

three coplanar forces are exerted on an object which is at rest. two of the forces are shown in the diagram. what is the magnitude of the third force if the object is to remain at rest? a 40.0 n b 56.6 n c 108 n d 120 n

Explanation:

Step1: Calculate the horizontal component of the 50N force

The horizontal component of the 50N force is \(50\cos53.1^{\circ}\). Using the value \(\cos53.1^{\circ}\approx0.6\), we get \(50\times0.6 = 30N\).

Step2: Calculate the vertical component of the 50N force

The vertical component of the 50N force is \(50\sin53.1^{\circ}\). Using the value \(\sin53.1^{\circ}\approx0.8\), we get \(50\times0.8=40N\).

Step3: Find the net horizontal force

The net horizontal force from the two given forces is \(70 - 30=40N\).

Step4: Find the magnitude of the resultant of the two given forces

The resultant force of the two given forces (using Pythagoras' theorem for the net horizontal \(40N\) and vertical \(40N\) components) is \(\sqrt{40^{2}+40^{2}}=\sqrt{1600 + 1600}=\sqrt{3200}\approx56.6N\). Since the object is in equilibrium, the third force must be equal in magnitude to this resultant.

Answer:

B. \(56.6N\)