Consider the figure below. (Let w1 = 150 N and w2 = 38.0 N.)
(a) What is the minimum force of friction required to hold the system of the figure in equilibrium?
N
(b) What coefficient of static friction between the 150-N block and the table ensures equilibrium?
(c) If the coefficient of kinetic friction between the 150-N block and the table is 0.127, what hanging weight should replace the 38.0-N weight to allow the system to move at a constant speed once it is set in motion?
N
To solve these problems, we first need to analyze the forces acting on both blocks. Let's call the 150-N block A and the 38.0-N weight B.
(a) The minimum force of friction required to hold the system in equilibrium is the force that balances the gravitational force acting on block B. So,
Friction force = w2 = 38.0 N
(b) The coefficient of static friction between the 150-N block and the table ensures equilibrium can be found by the formula:
Friction force = static friction coefficient * Normal force
Since the system is in equilibrium, the normal force on block A is equal to its weight, w1.
Static friction coefficient = Friction force / Normal force
Static friction coefficient = 38.0 N / 150 N = 0.2533
(c) To find the hanging weight that allows the system to move at a constant speed with a given coefficient of kinetic friction, we first find the kinetic friction force acting on the 150-N block:
Kinetic friction force = kinetic friction coefficient * Normal force
Kinetic friction force = 0.127 * 150 N = 19.05 N
Now, we need to find the weight that provides this amount of force as tension in the string. Since the system is moving at a constant speed, the tension in the string equals the kinetic friction force:
Tension = 19.05 N
This tension is also equal to the gravitational force acting on the hanging weight:
Hanging weight = Tension = 19.05 N
To find the minimum force of friction required to hold the system in equilibrium, we need to consider the forces acting on each object.
(a) The forces acting on the 150-N block are the weight (w1) and the normal force (N1) exerted by the table. The forces acting on the 38.0-N block are its weight (w2) and the normal force (N2) exerted by the table. The force of friction (f) will oppose the motion and prevent the system from moving:
Since the system is in equilibrium, the sum of all the forces acting on each block in the horizontal direction must be zero. This gives us the equation:
f - f2 = 0
The force of friction on the 150-N block is equal to the force of friction on the 38.0-N block. Therefore:
f = f2
Now, let's calculate the normal forces exerted on each block. The normal force N is equal in magnitude and opposite in direction to the weight of each block.
N1 = w1 = 150 N
N2 = w2 = 38.0 N
Since we want to find the minimum force of friction required, we need to consider the case where all the weight of block 2 is transferred to block 1. In this case, block 2 is essentially hanging freely, and its weight is fully supported by block 1.
So, the normal force N1 acting on block 1 will be the sum of the weights of both blocks:
N1' = N1 + N2 = 150 N + 38.0 N = 188 N
Now we can calculate the minimum force of friction:
f = f2 = μs * N1' = μs * 188 N
(b) To find the coefficient of static friction (μs) between the 150-N block and the table that ensures equilibrium, we can use the equation from part (a):
f = μs * N1' = μs * 188 N
To ensure equilibrium, the minimum force of friction f must be equal to the maximum static friction force:
f = μs * N1
Setting these two equations equal to each other and solving for μs:
μs * 188 N = 150 N
μs = 150 N / 188 N
μs ≈ 0.798
(c) To find the hanging weight that should replace the 38.0-N weight to allow the system to move at a constant speed once it is set in motion, we need to consider the forces acting on the system.
In this case, the force of friction will be kinetic, and the system will move at a constant speed. The equation for the force of kinetic friction is:
fk = μk * N1
Given the coefficient of kinetic friction (μk) as 0.127, we can substitute this value into the equation:
fk = 0.127 * N1
Since we want the system to move at a constant speed, the force required to overcome the kinetic friction is equal to the weight being hung:
fk = w
Substituting the values:
0.127 * N1 = w
Since N1 = N1' = 188 N (as calculated in part (a)), we can solve for the hanging weight w:
0.127 * 188 N = w
w ≈ 23.9 N
So, the hanging weight that should replace the 38.0-N weight to allow the system to move at a constant speed is approximately 23.9 N.
To answer these questions, we need to calculate the forces and use the conditions for equilibrium.
(a) The minimum force of friction required to hold the system in equilibrium can be found by summing the vertical forces acting on the system. In this case, the vertical forces are the weights of the two masses: w1 = 150 N and w2 = 38.0 N.
The total force acting downwards is given by F_total = w1 + w2 = 150 N + 38.0 N = 188.0 N.
Since the system is in equilibrium, the vertical forces must sum to zero. Therefore, the minimum force of friction required to hold the system in equilibrium is 188.0 N.
(b) The coefficient of static friction between the 150-N block and the table can be calculated using the equation for static friction, which is given by:
F_friction_static = coefficient_static_friction * F_normal
where F_normal is the force normal to the table's surface. In this case, F_normal is equal to the weight of the 150-N block, so F_normal = w1 = 150 N.
To find the coefficient of static friction, we rearrange the equation as follows:
coefficient_static_friction = F_friction_static / F_normal
Substituting the known values, we get:
coefficient_static_friction = 188.0 N / 150 N = 1.2533
Therefore, the coefficient of static friction between the 150-N block and the table that ensures equilibrium is approximately 1.2533.
(c) To find the hanging weight that should replace the 38.0-N weight to allow the system to move at a constant speed once it is set in motion, we need to consider the forces acting on the system when it is in motion.
When the system is in motion, the force of kinetic friction comes into play. The force of kinetic friction can be calculated using the equation:
F_friction_kinetic = coefficient_kinetic_friction * F_normal
Given that the coefficient of kinetic friction is 0.127, and the value of F_normal is still the weight of the 150-N block (150 N), we can calculate:
F_friction_kinetic = 0.127 * 150 N = 19.05 N
To maintain constant speed, the force pulling the system downwards (hanging weight) should balance the force of kinetic friction. Therefore, the hanging weight should be equal to the force of kinetic friction: 19.05 N.