Consider the 45 N weight held by two cables
shown below. The left-hand cable is horizon-
tal.
a) What is the tension in the cable slanted
at an angle of 50�?
Answer in units of N.
007 (part 2 of 2) 10.0 points
b)What is the tension in the horizontal cable?
well, if the right hand is horizontal, the left hand is holding the weight in the vertical
a) 45/tensionright = sin50 if the 50 degrees is measured from the horizontal.
The tension in the left cable has to equal the horizontalcomponent of tension in the slanted cable: tensionleft*cos50
To find the tension in the cable slanted at an angle of 50 degrees (part a):
1. Draw a free-body diagram of the weight and the cables.
2. Label the tension in the horizontal cable as T1 and the tension in the slanted cable as T2.
3. Split the weight into its vertical and horizontal components.
The weight's vertical component is given by Wv = W * sin(50), where W is the weight (45 N) and sin(50) is the sine of 50 degrees.
4. Set up an equilibrium equation for the vertical forces:
T2 + Wv = 0
5. Solve for T2:
T2 = - Wv = - (W * sin(50))
The tension in the slanted cable is - (W * sin(50)) N.
To find the tension in the horizontal cable (part b):
1. Set up an equilibrium equation for the horizontal forces:
T1 = W * cos(50)
The tension in the horizontal cable is W * cos(50) N.
Please note that the negative sign in the tension of the slanted cable indicates that the tension is acting in the opposite direction to the weight.
To find the tension in the cable slanted at an angle of 50 degrees (part a), and the tension in the horizontal cable (part b), we can use the concept of equilibrium and Newton's laws of motion.
Let's start with part a:
In order for the weight to remain in equilibrium (not accelerate), the net force acting on it must be zero. The forces acting on the weight are the tension forces in both cables.
We can break down the weight force into its components parallel and perpendicular to the slanted cable. The weight force has a magnitude of 45 N, and the angle between the weight force and the slanted cable is 50 degrees.
The component of the weight force perpendicular to the slanted cable (vertical direction) can be found using trigonometry:
F_perpendicular = Weight force * sin(angle) = 45 N * sin(50 degrees)
The tension force in the slanted cable must have the same magnitude but in the opposite direction to counterbalance this perpendicular component of the weight. Therefore:
Tension in the slanted cable = F_perpendicular = 45 N * sin(50 degrees)
Now let's move on to part b:
Since the weight is held in equilibrium, the net force in the horizontal direction must also be zero. The only force acting horizontally is the tension in the horizontal cable.
The component of the tension in the slanted cable parallel to the horizontal direction can be found using trigonometry:
F_parallel = Tension in the slanted cable * cos(angle) = [45 N * sin(50 degrees)] * cos(50 degrees)
Since the net force in the horizontal direction is zero, the tension in the horizontal cable must be equal in magnitude but in the opposite direction. Therefore:
Tension in the horizontal cable = - F_parallel = - [45 N * sin(50 degrees)] * cos(50 degrees)
By substituting the values of sin(50 degrees) and cos(50 degrees) into the above expressions, you can determine the numerical values for both tensions.
Remember to calculate the trigonometric functions using the appropriate units for angles (radians or degrees) as required by the question.