A block of mass M = 6.755 kg is pulled along a horizontal frictionless surface by a rope of mass m = 0.592 kg, as shown in Fig. 5-63. A horizontal force = 41.5 N is applied to one end of the rope. Assume that the rope only sags a negligible amount. Find (a) the acceleration of rope and block, (b) the force on the block from the rope, and (c) the tension in the rope at its midpoint.
(M)----m---->F
To solve this problem, we will use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.
First, let's calculate the acceleration of the rope and block system (a).
(a) The force applied to the rope (F) will create a tension force (T) in the rope and block system. We can calculate the tension by subtracting the weight of the rope (m * g) from the applied force (F). Since the system is on a frictionless surface, there is no normal force, only the force of gravity acting on each mass.
T = F - (m * g)
where g is the acceleration due to gravity (approximately 9.8 m/s^2).
Next, we need to calculate the net force acting on the system. The net force is the force on the block from the rope, which is equal to the tension force (T).
Force on the block = T
Finally, we can apply Newton's second law to find the acceleration of the system.
Net force = (M + m) * a
Since the force on the block is equal to the tension force (T), the equation becomes:
T = (M + m) * a
Now, let's solve for the unknowns:
(a) To find the acceleration of the rope and block system, rearrange the equation T = (M + m) * a to solve for a:
a = T / (M + m)
(b) The force on the block from the rope is equal to the tension force (T):
Force on the block = T
(c) The tension in the rope at its midpoint is also equal to the tension force (T).
Tension in the rope = T
Now, you can substitute the given values for the mass of the block (M), mass of the rope (m), and the applied force (F) into the equations to find the answers.