Two blocks lie on a triangle as shown in the figure. Block m has a mass of 5 kg and block P has a mass of 14 kg. Assume the inclines and pulleys are frictionless find the tension in the string.
They lie with the string resting on the peak of a 110 degree obtuse triangle, each side corner of the triangle measuring 35 degress
Ok, I am visulizing two inclines back to back, with a pulley at the top, and masses on each side.
on each block, the gravity component DOWN the plane is
mg*cos35
No resistive force (friction).
so first, find the motion.
Net force=(14-5)*9.8*cos35 twoward the greater mass.
net force=totamass*acceleration
acceleration=(14-5)*9.8*cos35 /19
Now knowing acceleration, one can get tension. Examine either block, here is the smaller block.
Tension-5*9.8*cos35=5*acceleration
solve for tension.
To find the tension in the string, we can analyze the forces acting on each block separately. Let's start with block P:
1. Draw a free-body diagram of block P, considering the forces acting on it.
- There will be two forces acting on block P: the weight (mg) and the tension in the string (T).
- The weight force (mg) is vertically downward, equal to the mass (14 kg) multiplied by the acceleration due to gravity (9.8 m/s^2).
2. Break down the weight force into its components.
- Since block P is on an inclined plane, we need to consider the components of the weight force parallel and perpendicular to the incline.
- The perpendicular component (mg * cos(35°)) opposes the normal force and cancels it out since the incline is frictionless.
- The parallel component (mg * sin(35°)) acts downward along the incline and needs to be considered in the following steps.
3. Apply Newton's second law of motion to determine the net force acting on block P in the direction of motion (parallel to the incline).
- The net force in the direction of motion is equal to the parallel component of the weight force minus the tension in the string.
- The net force can be written as: net force = (mg * sin(35°)) - T = m * a, where 'a' is the acceleration of the system (common for both blocks).
Now, let's move on to block m:
4. Again, draw a free-body diagram of block m.
- The weight force (mg) acts vertically downward, equal to the mass (5 kg) multiplied by the acceleration due to gravity (9.8 m/s^2).
- The tension in the string (T) acts upward, opposing the motion.
5. Apply Newton's second law of motion to determine the net force acting on block m.
- The net force is equal to T minus the weight force and can be written as: net force = T - (mg) = m * a.
6. Now, equate the net forces for both blocks.
- Since the masses of the blocks cancel each other out, we have the equation: (mg * sin(35°)) - T = T - (mg).
- Simplifying this equation, we get: 2T = (mg * sin(35°)) + (mg).
- Rearranging, we get: T = (mg * sin(35°)) / 2 + (mg) / 2.
Finally, we can substitute the given values to find the tension in the string:
T = (5 kg * 9.8 m/s^2 * sin(35°)) / 2 + (5 kg * 9.8 m/s^2) / 2.
T = 24.42 N + 24.5 N.
T ≈ 48.92 N.
Therefore, the tension in the string is approximately 48.92 N.