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Chapter 3: Problem 16

A fireman wants to slide down a rope. The breaking load for the rope is \(\frac{3}{4}\) th of the weight of the fireman. The acceleration of the fireman to prevent the rope from breaking will be (Acceleration due to gravity is \(g\) ) (A) \(g / 4\) (B) \(g / 2\) (C) \(3 g / 4\) (D) Zero

Short Answer

Expert verified
The acceleration of the fireman to prevent the rope from breaking is \(\frac{g}{4}\), which corresponds to answer choice (A).

Step by step solution

01

Identify the forces acting on the fireman.

There are two forces acting on the fireman: the gravitational force (weight) acting downward and the tension force in the rope acting upward.
02

Write the relation for the given breaking load limit of the rope.

The breaking load of the rope is \(\frac{3}{4}\) of the fireman's weight. If the fireman's mass is \(m\), then the weight is \(mg\). Therefore, the breaking load, which is the maximum tension allowed in the rope, is \(T_{max} = \frac{3}{4}mg\).
03

Apply Newton's second law of motion to the fireman.

According to Newton's second law of motion, the net force acting on the fireman is equal to the product of the mass and acceleration of the fireman, that is, \(F_{net} = ma\). The net force acting on the fireman is the difference between the tension force and the gravitational force acting on the fireman: \(F_{net} = T - mg\).
04

Calculate the acceleration that prevents rope from breaking.

To prevent the rope from breaking, we need to find an acceleration that makes the tension force equal to \(T_{max}\). Using the equations from Steps 3 and 4: \(ma = T - mg\) Since \(T = T_{max} = \frac{3}{4}mg\), we have: \(ma = \frac{3}{4}mg - mg\) Divide both sides of the equation by \(m\): \(a = \frac{3}{4}g - g\)
05

Simplify the expression for acceleration.

Subtract the acceleration due to gravity from the \(\frac{3}{4}\) multiple: \(a = \frac{3g - 4g}{4}\) \(a = -\frac{g}{4}\) Since we are looking for the magnitude of the acceleration, we can ignore the negative sign: \(a = \frac{g}{4}\) The acceleration of the fireman to prevent the rope from breaking is \(\frac{g}{4}\), which corresponds to answer choice (A).

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Most popular questions from this chapter

A body of mass \(1.5 \mathrm{~kg}\) is thrown vertically upwards with an initial velocity of \(40 \mathrm{~m} / \mathrm{s}\) reaches its highest point after \(3 \mathrm{~s}\). The air resistance acting on the body during the ascent is (assuming air resistance to be uniform, \(g=10 \mathrm{~m} / \mathrm{s}^{2}\) ) (A) \(35 \mathrm{~N}\) (B) \(25 \mathrm{~N}\) (C) \(15 \mathrm{~N}\) (D) \(5 \mathrm{~N}\)

One end of massless rope, which passes over a massless and frictionless pulley \(P\) is tied to a hook \(C\) while the other end is free. Maximum tension that the rope can bear is \(840 \mathrm{~N}\). With what value of maximum safe acceleration (in \(\mathrm{ms}^{-2}\) ) can a man of \(60 \mathrm{~kg}\) climb on the rope? (A) 16 (B) 6 (C) 4 (D) 8

A pebble of mass \(0.05 \mathrm{~kg}\) is thrown vertically upwards. The direction and magnitude of the net force on the pebble is given below, choose the incorrect option. (A) During its upward motion, force is \(0.5 \mathrm{~N}\) in vertically upward. (B) During its downward motion, force is \(0.5 \mathrm{~N}\) in vertically downward. (C) At the highest point, where it is momentarily at rest, force is \(0.5 \mathrm{~N}\) in vertically downward. (D) If the pebble was thrown at an angle of say \(45^{\circ}\) with the horizontal direction, force is \(0.5 \mathrm{~N}\) in vertically downward (Ignoring air resistance).

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