Resolving the forces for an object suspended on two strings.

Imagine a following situation: a stationary (not moving) object with a mass of 5 kg is suspended in the air by two strings. The angles between strings and the vertical directions are 15 and 55 degrees. What are tensions T1 and T2 in both strings?The first thing to do would be just drawing a simple diagram. I don't think it's possible to add one here, but I drew a little diagram myself! Now, we have to resolve the forces in the horizontal and vertical directions. Because the object is stationary, according to Newton's laws of dynamics we know that adding all the forces will equal 0. Let's start!Vertical forces: we have two vertical components of tension T1 and T2 pointing upwards (call it +ve direction) and weight i.e. 5g pointing downwards (let's take g = 9.8). So: T1cos(15o) + T2cos(55o) - 59.8 = 0  (eqn. 1)Horizontal forces: one component points to the left (-ve direction), the other to the right (+ve direction). Hence: T1sin(15o) - T2sin(55o) = 0  (eqn. 2)We can see equations 1 and 2 form a set of simultaneous equations. Let's rearrange 2 to get T1 in terms of T2: T1 = T2sin(55o) / sin(15o)  (eqn. 3).Substituting equation 3 into equation 1 gives: T2sin(55o)*cot(15o) + T2cos(55o) = 49  (eqn. 4) .From equation 4: T2 = 13.5 N and hence from equation 3: T1 = 42.7 N.

FW
Answered by Filip W. Physics tutor

18340 Views

See similar Physics A Level tutors

Related Physics A Level answers

All answers ▸

How can you tell if a reaction will happen?


Describe the photoelectric effect and what it tells us about the properties of light .


What is damping in Simple Harmonic Motion?


Initially, trucks A and B are travelling in opposite directions. A has mass 1000 kg and is travelling at speed 7ms^-1. B has mass 4000kg and is travelling at speed 2ms^-1. What is their speed and direction after collision if they move together?


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

MyTutor is part of the IXL family of brands:

© 2025 by IXL Learning