A 4.95 kg box is suspended from a system of massless wires...

Name: A 4.95 kg box is suspended from a system of massless wires attached to a ceiling. The first wire makes an angle of ?1 = 58.6° with the ceiling, whereas the second wire makes an angle of ?2 = 44.5° with the ceiling. Determine the magnitudes of the tensions in the first and second wires, T1 and T2, respectively. T1 = N T2 = N
Uploaded: 2023-08-16T09:29:57-08:00
Duration: 5 min
Channel: Madhur L
Description: A 4.95 kg box is suspended from a system of massless wires attached to a ceiling. The first wire makes an angle of ?1 = 58.6° with the ceiling, whereas the second wire makes an angle of ?2 = 44.5° with the ceiling. Determine the magnitudes of the tensions in the first and second wires, T1 and T2, respectively. T1 = N T2 = N

A 4.95 kg box is suspended from a system of massless wires attached to a ceiling. The first wire makes an angle of ?1 = 58.6° with the ceiling, whereas the second wire makes an angle of ?2 = 44.5° with the ceiling. Determine the magnitudes of the tensions in the first and second wires, T1 and T2, respectively. T1 = N T2 = N

Transcript

00:01 Hello students, in this question two wires are attached and it is making angle ?1 and ?2 with the horizontal tension is acting on both wire that is t1 and t2 and a box of mass 4 .95 kg it is attached to the another wire and which is having a tension t3.

00:22 So, we can resolve the tension into its horizontal and vertical components.

00:28 So, the horizontal component of both tension will be t1 cos ?1 and t2 cos ?2 and the vertical components of both tensions t1 and t2 will be t1 sin ?1 and t2 sin ?2 and there is a force of gravity that is acting downwards mg that is equal to mg and there is a tension t3 that is acting upwards and this t3 will be equal to mg.

00:53 So, since the system is in equilibrium the total force acting on the horizontal and vertical direction will be 0.

01:00 So, first we can take the horizontal direction horizontal direction so in the horizontal direction the total force f total will be equal to 0.

01:19 So, t1 cos ?1 minus t2 cos ?2 is equal to 0.

01:32 So, t1 by t2 will be equal to cos ?2 divided by cos ?1.

01:41 So, this will be equal to cos 44 .5 degree divided by cos 58 .6 degree.

01:57 So, from this we will get t1 will be equal to 1 .368 t2.

02:05 So, we can give this as equation number 1.

02:09 Next we can take the total force acting on the vertical direction.

02:14 So, in the vertical direction so in the vertical direction total force acting will be 0 since the system is in equilibrium.

02:28 So, this will be t3 t3 minus t1 sin ?1 minus t2 sin ?2 equal to 0.

02:47 So, here tension t3 is equal to mg...