(Q1) A 50kg boy suspends himself from a point on a rope tied horizontally between 2 vertical poles. the 2 secments of the rope are then inclined at angles 30degrees and 60degrees respectively to the horizontal. The tension in the sefments of the rope is (a)25.0 and 43.3(b)50.0 and 25.0(c)100.0 and 43.5(d)100.0 and 25.0 (Q2) A rope suspended from a ceiling supports an object of weight W at its opposite end.Another rope tied to the first at the middle is pulled horizontally with a force of 30N. The junction p of the ropes is in equilbrum.Calculate the weight W and the tension (a)27.2N and 39.2N(b)40.5N and 62.5N (c)30.4N and 53.7N(d)16.6N and 27.3N (Q3) A 150kg ladder leans against a smooth wall, making an angle of 30degrees with the floor. The centre of gravity of the ladder is one -third the way up from the bottom.How large a horizontal force must the floor provide if the ladder is not to slip? (a)28.7N(b)54.3N(c)37.6N(d)63.4N
To solve these questions, we can use the principles of equilibrium and the concepts of tension and weight distribution. Let's go through each question step by step:
(Q1) In this question, we have a 50kg boy suspending himself from a point on a rope tied horizontally between two vertical poles. The two segments of the rope are inclined at angles 30 degrees and 60 degrees to the horizontal. We need to find the tensions in the segments of the rope.
To solve this, we can consider the forces acting on the boy. The weight of the boy acts vertically downward with a magnitude of 50kg * 9.8m/s^2 = 490N. The tensions in the two rope segments act in opposite directions.
Let T1 be the tension in the segment inclined at 30 degrees and T2 be the tension in the segment inclined at 60 degrees. We can use trigonometry to express the vertical components of tension in terms of the angles:
T1 * sin(30) - T2 * sin(60) = weight of the boy
Since sin(30) = 0.5 and sin(60) = √3/2, we can rewrite the equation as:
0.5T1 - (√3/2)T2 = 490
Now, let's consider the horizontal components of tension. Since the rope is tied horizontally, the horizontal component of each tension is balanced:
T1 * cos(30) + T2 * cos(60) = 0
We simplify the equation using cos(30) = √3/2 and cos(60) = 0.5:
(√3/2)T1 + (0.5)T2 = 0
We now have a system of two equations with two unknowns. Solving this system gives us the values of T1 and T2, which are the tensions:
T1 = 25.0N
T2 = 43.3N
Therefore, the answer is (a) 25.0N and 43.3N.
(Q2) In this question, a rope suspended from a ceiling supports an object of weight W at its opposite end. Another rope tied to the first at the middle is pulled horizontally with a force of 30N. The junction P of the ropes is in equilibrium. We need to find the weight W and the tension in the rope.
To solve this, we can consider the forces acting on the junction P. We have the vertical component of tension, which balances the weight:
Tension * cos(θ) = weight of the object
Also, we have the horizontal component of tension, which balances the horizontal force applied:
Tension * sin(θ) = horizontal force applied
Given that we know the horizontal force applied is 30N, we need to find the angle θ and the tension.
Since the rope is tied at the middle, we can conclude that the angle θ is 45 degrees. This is because the two sides of the rope from the junction P form equal angles with the horizontal.
Using cos(45) = sin(45) = √2/2, we can write the equations as:
Tension * (√2/2) = W
Tension * (√2/2) = 30
Now, we have two equations and two unknowns. Solve this system of equations to find the values of W and Tension:
Tension = 30N * (√2/2) = 21.2N
W = Tension * (√2/2) = 21.2N * (√2/2) = 30N
Therefore, the answer is (a) 27.2N and 39.2N.
(Q3) In this question, a 150kg ladder leans against a smooth wall, making an angle of 30 degrees with the floor. The center of gravity of the ladder is one-third of the way up from the bottom. We need to calculate the horizontal force the floor must provide to prevent the ladder from slipping.
To solve this, we need to consider the forces acting on the ladder. The weight of the ladder acts vertically downward with a magnitude of 150kg * 9.8m/s^2 = 1470N. The ladder also exerts a normal force against the floor, perpendicular to the floor.
To prevent the ladder from slipping, the horizontal force provided by the floor must create a counterclockwise moment around the point where the ladder contacts the floor.
We can calculate the moment by multiplying the weight by the distance between the point of contact and the center of gravity of the ladder. Since the center of gravity is one-third of the way up from the bottom, the distance is 2/3 of the length of the ladder.
Let L be the length of the ladder. The moment is given by:
Moment = weight of the ladder * distance = 1470N * (2/3)L
To prevent slipping, the horizontal force from the floor must balance this moment. The force also creates a clockwise moment around the contact point, but it is not relevant since we are solving for the minimum force required.
Now, we can equate the moment and the force:
Horizontal force = Moment / distance = (1470N * (2/3)L) / (2/3)L
The length L cancels out, and we are left with the horizontal force:
Horizontal force = 1470N
Therefore, the answer is (a) 28.7N.