"Two blocks lie on a triangle as shown in the figure. Block m has a mass of 8 kg and block P has a mass of 21 kg. Assume the inclines and pulleys are frictionless find the tension in the string." The two blocks are on opposite side of an isosceles triangle with the two lower angles of the triangle measuring 35 degrees each. The blocks hang from a pulley sticking out of the top of the triangle.
It matters what the angles are.
To find the tension in the string, we can start by resolving the forces acting on each block separately.
Let's begin with block m. It is on an incline, so we need to consider the forces acting along the incline and perpendicular to it.
The force acting along the incline is the component of its weight parallel to the incline, which we can calculate as mg*sin(35), where m is the mass of block m and g is the acceleration due to gravity.
Considering the forces perpendicular to the incline, we only have the normal force, which is equal to the component of its weight perpendicular to the incline. This can be calculated as mg*cos(35).
Now let's move on to block P. Since it is hanging vertically, the forces acting on it are its weight, which is equal to mg, and the tension in the string.
Next, we can set up equations based on Newton's second law for each block:
For block m:
mg*sin(35) - T = ma, where a is the acceleration of block m.
For block P:
T - mg = ma, where a is the acceleration of block P.
Since the two blocks are connected by a string and have the same acceleration, we can set their equations equal to each other:
mg*sin(35) - T = T - mg.
Now we can solve this equation to find the tension in the string.
Rearranging the equation, we have:
2T = mg*sin(35) + mg.
Dividing both sides by 2, we get:
T = (mg*sin(35) + mg) / 2.
Substituting the given values of m and g, we get:
T = (8 * 9.8 * sin(35) + 8 * 9.8) / 2.
Evaluating this expression will give us the tension in the string.