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Chapter 3: Problem 149

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

Short Answer

Expert verified
The maximum safe acceleration with which the man can climb the rope without exceeding the maximum tension is approximately 4 \(ms^{-2}\).

Step by step solution

01

Analyze the forces acting on the man while climbing

When the man is climbing the rope, his weight will be acting downwards, and the tension in the rope will be acting upwards. We can represent this by drawing a free body diagram for the man. We can also represent the force of tension acting on the man, due to the massless rope. We will assume that the man is climbing up the rope with an acceleration "a."
02

Setting up the equation

Using Newton's second law of motion, we can write an equation for the forces acting on the man. \(T - mg = ma\) Where: T is the tension in the rope (in Newtons) m is the mass of the man (60 kg) g is the acceleration due to gravity (\(9.8 ms^{-2}\)) a is the acceleration of the man while climbing (in \(ms^{-2}\))
03

Solve the equation for acceleration "a"

As we need to find the maximum safe acceleration, we need to ensure that the tension in the rope should not exceed the maximum tension it can bear (840 N). Maximum tension in the rope (T) should be equal to 840 N. Therefore, \(840 - 60 * 9.8 = 60 * a\)
04

Calculate the maximum safe acceleration "a"

Now, we will simply solve the equation for "a": \(a = \frac{840 - (60 * 9.8)}{60}\) \(a = \frac{840 - 588}{60}\) \(a = \frac{252}{60}\) \(a = 4.2 ms^{-2}\) Since 4.2 is very close to 4, we can conclude that the maximum safe acceleration with which the man can climb the rope without exceeding the maximum tension is approximately 4 \(ms^{-2}\). Therefore, the correct answer is option (C).

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Most popular questions from this chapter

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).

A light spring balance hangs from the hook of the other light spring balance and a block of mass \(M \mathrm{~kg}\) hangs from the former one. Then the true statement about the scale reading is (A) Both the scales read \(M \mathrm{~kg}\) each (B) The scale of the lower one reads \(M \mathrm{~kg}\) and of the upper one zero (C) The reading of the two scales can be anything but the sum of the readings will be \(M \mathrm{~kg}\) (D) Both the scales read \(M / 2 \mathrm{~kg}\)

The driver of three-wheeler moving with a speed of \(36 \mathrm{~km} / \mathrm{h}\) sees a child standing in the middle of the road and brings his vehicle to rest in \(4.0 \mathrm{~s}\) just in time to save the child. What is the average retarding force on the vehicle? The mass of the three-wheeler is \(400 \mathrm{~kg}\) and the mass of the driver is \(65 \mathrm{~kg}\). (A) \(1162.5 \mathrm{~N}\) (B) \(116.25 \mathrm{~N}\) (C) \(1112 \mathrm{~N}\) (D) None of these

A block is kept on a frictionless inclined surface with angle of inclination \(\alpha\). The incline is given an acceleration a to keep the block stationary with respect to the incline. Then a is equal to (A) \(g / \tan \alpha\) (B) \(g \operatorname{cosec} \alpha\) (C) \(g\) (D) \(g \tan \alpha\)

A car of mass \(m\) starts from rest and acquires a velocity along east, \(v=v \hat{i}(v>0)\) in two seconds. Assuming the car moves with uniform acceleration, the force exerted on the car is (A) \(\frac{m v}{2}\) eastward and is exerted by the car engine. (B) \(\frac{m v}{2}\) eastward and is due to the friction on the tyres exerted by the road. (C) more than \(\frac{m v}{2}\) eastward exerted due to the engine and overcomes the friction of the road. (D) \(\frac{m v}{\text { exerted by the engine. }}\)
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