a framed picture of weight 15N is to be hung on a wall using a price of string . the end of the string are tied to two points ,0.60m apart on the same horizontal level ,on the back of the picture .Find the tension in the string if the string is (a) 1.0 m long
(b)0.66m long
+2 first year science
a. Cos A = 0.3/0.5 = 0.6.
A = 53.13o = B.
T1*sin53.13 + T2*sin(180-53.13) = +15.
T2 = T1:
T1*sin53.13 + T1*sin126.9 = 15
0.8T1 + 0.8T1 = 15
T1 = 9.375 N. = Tens1on in the string.
b. Cos A = 0.3/0.33 = 0.90909,
A = 24.62o.
T1*sin24.62 + T2*sin(180-24.62) = +15.
T2 = T1
T1 = ?.
Use the same procedure as part "a" to finish the problem.
Its okay
It is Awesome
(a) Well, if the string is 1.0 m long, it means it will be hanging perfectly vertically from the wall. So, the tension in the string would simply be the weight of the picture, which is 15N. Easy peasy!
(b) Ah, now we have a little angle involved. With a string length of 0.66m, it means the string will be at an angle with the wall. Now, to find the tension, we need to use some trigonometry.
We can consider this a right triangle, where the hypotenuse is the length of the string (0.66m), and the adjacent side is the horizontal distance between the two attachment points (0.60m).
So, using the cosine function, we can find the tension:
cosθ = adjacent/hypotenuse
cosθ = 0.60m/0.66m
Now, let's solve for θ:
θ = cos^(-1)(0.60m/0.66m)
Finally, we can find the tension by dividing the weight of the picture (15N) by the cosine of θ:
Tension = 15N/cosθ
And there you have it, the tension in the string! Now you can hang that picture and admire it without any worries. Good luck!
To find the tension in the string, we can start by drawing a free body diagram for the framed picture.
First, we need to identify the forces acting on the framed picture:
1. Weight (W) of the picture, acting downward (15N).
2. Tension forces (T) in the string, acting horizontally towards the points where the string is tied.
Let's calculate the tension in the string for both scenarios:
(a) When the string is 1.0m long:
To find the tension, we need to analyze the forces in the horizontal direction. Since the picture is in equilibrium (not moving), the tension force on each side of the string will be equal.
Let's assume T1 is the tension on the left side of the string, and T2 is the tension on the right side of the string.
Using the equation for equilibrium in the horizontal direction:
T1 + T2 = 0
Considering that the distance between the points where the string is tied is 0.60m, the following relationship holds:
T2 = T1 + 0.60N
Now, let's consider the vertical forces. The downward force of the weight (15N) will result in a vertical component of tension.
Using trigonometry, we can find the vertical component of tension:
T_vertical = 15N * sin(θ)
θ is the angle between the weight vector and the vertical axis.
Since the picture is hanging, the vertical component of tension must balance the weight. Therefore:
T_vertical = 15N
The tension force in the string can be determined using Pythagoras' theorem:
T^2 = T_horizontal^2 + T_vertical^2
Substituting the known values:
T^2 = (T1 + T2)^2 + (15N)^2
With the relationship between T1 and T2:
T^2 = (T1 + T1 + 0.60N)^2 + (15N)^2
Now we can solve for T1:
T^2 = (2T1 + 0.60N)^2 + (15N)^2
T1 = (sqrt(T^2 - (15N)^2) - 0.60N) / 2
Substituting T1 into the equation for T2:
T2 = T1 + 0.60N
(b) When the string is 0.66m long:
The method to calculate the tension in the string is the same as in case (a), but this time using the length of 0.66m instead of 1.0m. Repeat the calculations using this new length.
Note: In both scenarios, make sure to consider the appropriate significant figures and units throughout the calculations.