Two blocks of mass m1 = 7 kg and m2 = 6 kg connected with a light in-extensible string are placed on a horizontal table. A third block of mass m=3 kg is attached with a string to block 2 over a pulley and hangs off the edge of the table. The coefficient of friction between the first block and the table is µ1 = 0.2, between the second block and the table is µ2 = 0.1. Find the tension in both strings and the friction forces between the table and both blocks
g =9.81 m/s^2
F1= friction force on m1 = 0.2 * 7 g = 1.4 g
F2 = friction force on m2 = 0.1 * 6 g = 0.6 g
F3 = gravity force down on m3 = 3 g
T2 = tension on blocks 2 and 3
for block3:
3 g - T2 = 3 a
T1 = tension on blocks 1 and 2
block2:
T2 - T1 - 0.6 g = 6 a
block 1
T2 - 1.4 g = 7 a
so three equations and three unknowns T1, T2, a ,(remember g = 9.81)
T2 - 1.4 g = 7 a
T2 - T1 - 0.6 g = 6 a
3 g - T2 = 3 a
thankss! can I just solve for the unknowns simultaneously?
Precisely
To find the tension in both strings and the friction forces between the table and both blocks, we need to analyze the forces acting on each block.
Let's start with Block 1 (m1 = 7 kg). The forces acting on Block 1 are:
1. Weight (mg1): The weight of Block 1 acts vertically downward with a magnitude of mg1, where g is the acceleration due to gravity (9.8 m/s^2).
2. Normal force (N1): The table exerts an upward force on Block 1, perpendicular to the table surface.
3. Friction force (f1): The friction force acts horizontally opposite to the direction of motion or impending motion between Block 1 and the table. The magnitude of this force is given by f1 = µ1 * N1, where µ1 is the coefficient of friction between Block 1 and the table.
The tension in the string connecting Block 1 and Block 2 will be the same (T) because the string is light and inextensible. So, the magnitude of the tension is T.
Now, let's move on to Block 2 (m2 = 6 kg). The forces acting on Block 2 are:
1. Weight (mg2): The weight of Block 2 acts vertically downward with a magnitude of mg2.
2. Tension (T): The tension in the string connecting Block 1 and Block 2 acts horizontally, pulling Block 2 towards the right.
3. Normal force (N2): The table exerts an upward force on Block 2, perpendicular to the table surface.
4. Friction force (f2): The friction force acts horizontally opposite to the direction of motion or impending motion between Block 2 and the table. The magnitude of this force is given by f2 = µ2 * N2, where µ2 is the coefficient of friction between Block 2 and the table.
Now, let's consider Block 3 (m3 = 3 kg) hanging off the edge of the table. The only force acting on Block 3 is its weight (mg3), which acts vertically downward.
To find the tension in the string connecting Block 2 and Block 3, we can use Newton's second law of motion. We know that the net force on Block 2 is equal to its mass times acceleration (F = ma). In this case, the net force on Block 2 is given by the tension (T) minus the friction force (f2).
So, F2 = T - f2 = m2 * a, where a is the acceleration of Block 2.
To find the acceleration, we can consider the system formed by Block 2 and Block 3. The net force on this system is the difference between the weight of Block 2 (mg2) and the weight of Block 3 (mg3), since T will get canceled out. So,
F_net = mg2 - mg3 = (m2 - m3) * a
Simplifying the equation, we get:
(T - f2)/m2 = (m2 - m3) * a
We can now substitute the known values into this equation to solve for the tension (T).
Similarly, we can find the friction forces (f1 and f2) by substituting the known coefficients of friction (µ1 and µ2) and using the equations f1 = µ1 * N1 and f2 = µ2 * N2.
In summary, to find the tension in both strings and the friction forces between the table and both blocks, we need to analyze the forces acting on each block and use Newton's second law of motion.