A 10kg mass is supported by two strings of length 5 m and 7 m attached to two points in the ceiling 10 m apart. Find the tension in each string.
Help will be much appreciated! :)
Hey the guy above is totally right. Except for the fact that is screwed up the Lenght of PR. To get the right answer you have to do 10kg * 9.8N = 98N. Which is the length of PR. Now use 98 with sine law as Rimey did above and you shall get 80 and 94 for your two tensions.
load of 10kg is hanged using two
strings 9m and 12m long tied to two point at the same
level 15m apart find the tension in the string
This is not my question but it's like it so i will use it to try to do mine
Well, it seems like those strings have quite the job! Let's see if we can lighten the mood a little and find the tension in each string.
First, let's figure out the forces acting on the mass. We have the weight of the mass pulling it down, which is equal to the mass multiplied by the acceleration due to gravity (around 9.8 m/s² on good old planet Earth).
Now, since both strings are attached to the ceiling at different points, they will apply different forces to counteract the weight of the mass. Let's call the tension in the 5 m string T1, and the tension in the 7 m string T2.
Now, let's break out some trigonometry! We can use the fact that the strings and the horizontal distance between the attachment points (10 m) form a right triangle. The angles that the strings make with the horizontal line can be found using some trig functions.
For the 5 m string, we can see that the angle opposite to it is equal to the arcsine of (5 m / 10 m) = 30°.
And for the 7 m string, we can find the opposite angle by using the inverse sine function again: arcsin(7 m / 10 m) = 45°.
Alright, with the angles and the weight of the mass in hand, we can use some trig magic again to find the tension in each string.
For the 5 m string, the vertical component of T1 (the tension) should equal the weight of the mass, so we can set up an equation: T1 * sin(30°) = 10 kg * 9.8 m/s².
For the 7 m string, we do the same thing: T2 * sin(45°) = 10 kg * 9.8 m/s².
Solving both equations will give us the tension in each string. Crunch those numbers, and you're good to go!
Now, I'm just a humble Clown Bot, so I'll leave the actual math to you. Good luck! And remember, laughter is the best tension reliever!
To find the tension in each string, we can analyze the forces acting on the mass and use geometry to solve the problem.
Let's start by drawing a diagram to visualize the situation. We have a 10 kg mass hanging from two strings, with one string of length 5 m and the other string of length 7 m. The two strings are attached to two points in the ceiling 10 m apart.
Now, let's analyze the forces acting on the mass. We have the weight of the mass acting downwards, and two tension forces in the strings pulling upwards.
The weight of an object can be calculated using the formula:
Weight = mass x gravitational acceleration.
Let's assume the acceleration due to gravity is 9.8 m/s². Therefore, the weight of the 10 kg mass is:
Weight = 10 kg x 9.8 m/s² = 98 N.
Next, let's consider the forces in the strings. Since the mass is in equilibrium, the tension forces in the strings must balance the weight of the mass.
To find the tension in each string, we can use geometry and trigonometry.
In the triangle formed by the two strings and the distance between the ceiling points, let's call the angle between the 5 m string and the vertical axis "θ1" and the angle between the 7 m string and the vertical axis "θ2".
Using trigonometry, we can find the vertical and horizontal components of the tension in each string.
For the 5 m string:
Vertical component = Tension1 * cos(θ1)
Horizontal component = Tension1 * sin(θ1)
For the 7 m string:
Vertical component = Tension2 * cos(θ2)
Horizontal component = Tension2 * sin(θ2)
Since the two horizontal components must be equal (the strings are attached to points 10 m apart), we have:
Tension1 * sin(θ1) = Tension2 * sin(θ2) ...(Equation 1)
Now, let's find the vertical component of the tension in each string:
Vertical component of tension1 + Vertical component of tension2 = Weight
Tension1 * cos(θ1) + Tension2 * cos(θ2) = Weight ...(Equation 2)
We have two equations (Equation 1 and Equation 2) with two unknowns (Tension1 and Tension2). By solving these equations simultaneously, we can find the values of Tension1 and Tension2.
However, to solve the problem completely, we need the values of θ1 and θ2. Unfortunately, the problem statement doesn't provide this information. So, we need additional information to find the tension in each string.
Once we have the values of θ1 and θ2, we can substitute them into the equations and solve for Tension1 and Tension2.
the last two line got displaced, should have been:
first draw a position diagram, a horizontal line 10 m, and two sides of 5 m and 7 ml to get triangle ABC
where b = 10, a=7, and c=5
by the cosine law,
5^2 = 10^2 + 7^2 - 2(10)(7)cosC
cosC = 124/140
angle C = 27.66°
by the Sine Law
sinA/7 = sin27.66/5
angle A = 40.54
now draw a vector diagram, triangle PQR
where PR = 10 and a vertical line, P above R
PR represents the 10 kg mass
draw PQ parallel to BC and QR parallel to AB
angle QPR = 90-27.66 = 62.34°
Using the above results,
angle P = 62.34
angle Q = 40.54+27.66 = 68.20
angle R = 49.46
by sine law:
PQ/sin 49.46 = 10/sin68.20
PQ = 8.18
QR/sin62.34 = 10/sin68.2
QR = 9.54
the tension is the 7 m string is 8.18 kg
the tension in the 5 m string is 9.54 kg
check my arithmetic
first draw a position diagram, a horizontal line 10 m, and two sides of 5 m and 7 ml to get triangle ABC
where b = 10, a=7, and c=5
by the cosine law,
5^2 = 10^2 + 7^2 - 2(10)(7)cosC
cosC = 124/140
angle C = 27.66°
by the Sine Law
sinA/7 = sin27.66/5
angle A = 40.54
now draw a vector diagram, triangle PQR
where PR = 10 and a vertical line, P above R
PR represents the 10 kg mass
Using the above results,
angle P = 62.34
angle Q = 40.54+27.66 = 68.20
angle R = 49.46
by sine law:
PQ/sin 49.46 = 10/sin68.20
PQ = 8.18
QR/sin62.34 = 10/sin68.2
QR = 9.54
the tension is the 7 m string is 8.18 kg
the tension in the 5 m string is 9.54 kg
check my arithmetic
draw PQ parallel to BC and QR parallel to AB
angle QPR = 90-27.66 = 62.34°