2.1 Two blocks of mass M kg and 2.5kg respectively are connected by a light, inextensible string. The string runs over a light, frictionless pulley ,The blocks are stationary
2.1.1 calculate the tension in a string
The coefficient of static friction between the unknown mass M and the surface of the table is 0.2.
2.1.2 Calculate the minimum value of M that'll prevent the blocks from moving
Fnet=ma
T-fs=ma
24.5-fs=M (0)
-fs=-24.5
:fs=24.5N
Fs=0.2×(M)(9.8)
24.5=1.96M
M=24.5/1.96
M=12.5kg
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Fnet=0
T1-Fg=0
T1=mg
T1=2,5(9,8)
T1 =24,5
Fnet =0
T1+(-fk)=0
24,5-0,2N=0
N=122,5
Fn=m(9,8)
122,5÷9,8=9,8m÷9,8
M=12,5
To calculate the tension in the string, we need to consider the forces acting on each block. Let's denote the tension in the string as T.
For the block with mass M kg, the only force acting on it is the tension in the string pulling it upwards. Since the block is stationary, the tension in the string must be equal to the downward force of gravity acting on the block. Therefore, we have:
T = M * g (Equation 1)
Where g is the acceleration due to gravity (approximately 9.8 m/s^2).
For the block with mass 2.5 kg, there are two forces acting on it. The tension in the string pulls it upwards, and the force of gravity pulls it downwards. Since the block is stationary, these two forces must balance each other. Therefore, we have:
T = 2.5 * g (Equation 2)
Now, let's calculate the minimum value of M that will prevent the blocks from moving. To do this, we need to consider the limiting condition, which is when the static friction between the unknown mass M and the surface of the table is at its maximum value.
The maximum static friction force can be calculated using the equation:
F_max = μ_s * N
Where μ_s is the coefficient of static friction and N is the normal force on the block (which is equal to the force of gravity acting on the block). In this case, the normal force N is given by M * g.
Therefore, the maximum static friction force is:
F_max = μ_s * M * g
For the blocks to be stationary, the tension in the string T must be equal to or less than the maximum static friction force F_max. So, we have:
T ≤ F_max
Substituting the values of T (from Equation 1) and F_max, we get:
M * g ≤ μ_s * M * g
Canceling out the common factors of g and rearranging the equation, we find:
M ≤ μ_s * M
Dividing both sides by μ_s, we have:
M/μ_s ≤ M
Since μ_s is given as 0.2, we can calculate the minimum value of M that prevents the blocks from moving by plugging in this value:
M/0.2 ≤ M
Simplifying the inequality, we get:
M ≤ 5M
Dividing both sides by M, we get:
1 ≤ 5
Since 1 is less than 5, the inequality is satisfied. Therefore, there is no specific minimum value of M that prevents the blocks from moving. Any value of M will keep the blocks stationary as long as the tension in the string does not exceed the maximum static friction force.
M=2.5 /0.2=12.5 kg
I need the answers and steps
tension = 2.5 g = 2.5 * 9.81 Newtons
0.2 M g = 2.5 g
M = 2.5 / 0.2 = 12.5 kg