A picture hangs on the wall suspended by two strings, as shown in the
figure, with θ = 65°. The tension in string 1 is 1.3 N.
(a)Is the tension in string 2 greater than, less than, or equal to the tension in string 1? Explain.
(b) Verify your answer to part (a) by calculating the tension in string 2
(c) What is the weight of the picture?
A baseball player slides into third base with an initial speed of 7.75 m/s. If the coefficient of kinetic friction between the player and the ground is 0.45, how far does the player slide before coming to rest?
1. Are the two strings tied to the same point on the wall?
2. a = u*g = 0.45 * (-9.8)=-4.41 m/s^2.
V^2 = Vo^2 + 2a*d.
V = 0.
Vo = 7.75 m/s.
a = 4.41 m/s^2.
Solve for d.
1a. 1.3*Cos(180-65) = -T2*Cos65.
-0.549 = -0.423T2.
T2 = 1.30 N.
The Tensions are equal because the angles are equal.
To analyze the situation and answer the questions, we need to understand the forces acting on the picture and how they relate to each other. Let's begin with the free-body diagram of the picture.
1) The tension in string 1 is given as 1.3 N. We'll use this information to analyze the forces acting on the picture.
(a) To determine whether the tension in string 2 is greater than, less than, or equal to the tension in string 1, we need to consider the equilibrium of forces acting on the picture.
In this case, there are essentially three forces acting on the picture:
- The tension force in string 1 (T1) acting at an angle of 65°.
- The tension force in string 2 (T2) acting at an angle of 65° but in the opposite direction.
- The weight of the picture (W) acting vertically downwards.
In equilibrium, the sum of the forces in the horizontal direction and the sum of the forces in the vertical direction must both be zero.
Considering the horizontal forces, we can resolve them into components:
- T1 * cos(65°) acting to the right.
- T2 * cos(65°) acting to the left.
Since the picture is in equilibrium, the horizontal forces must cancel each other out:
T1 * cos(65°) = T2 * cos(65°).
Dividing both sides of the equation by cos(65°), we find that:
T1 = T2.
Therefore, the tension in string 2 (T2) is equal to the tension in string 1 (T1).
(b) Now let's verify our answer by calculating the tension in string 2.
Since we already know that T1 = T2, we can calculate T2 by substituting the given value of T1.
T1 = T2 = 1.3 N.
So, the tension in string 2 is 1.3 N.
(c) Finally, to determine the weight of the picture (W), we need to consider the vertical forces acting on the picture.
The vertical forces are:
- T1 * sin(65°) acting upwards.
- T2 * sin(65°) acting downwards.
- The weight of the picture (W) acting downwards.
Since the picture is in equilibrium, the vertical forces must also cancel each other out:
T1 * sin(65°) + T2 * sin(65°) = W.
Substituting the values of T1 and T2, we get:
1.3 N * sin(65°) + 1.3 N * sin(65°) = W.
Simplifying the equation, we find that:
W = 2 * 1.3 N * sin(65°).
Now, you can calculate the weight of the picture using this equation.
Ti*sin(180-65) + T2*sin65 = -Wp*sin270.
1.3*0.90631 + 1.3*90,631 = Wp.
Wp = 2.36 N. = Wt. of picture.