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Chapter 3: Problem 48

A lift is moving downwards with an acceleration equal to acceleration due to gravity. A body of mass \(M\) kept on the floor of the lift is pulled horizontally. If the co-efficient of friction is \(\mu\), then the frictional resistance offered by the body is (A) \(M g\) (B) \(\mu \mathrm{Mg}\) (C) \(2 \mu \mathrm{Mg}\) (D) Zero

Short Answer

Expert verified
The frictional resistance offered by the body is (D) Zero.

Step by step solution

01

Analyze Vertical Forces

Since the body of mass M is kept on the floor of the lift, there are two forces acting vertically on the body. Firstly, there's the weight of the body acting downward, which is given by \(M*g\). Secondly, there's a normal force acting perpendicular to the floor of the lift and pointing upward, we will call it (N). Since the lift is moving downwards with an acceleration (g), body M's acceleration is also (g) downward. By using the second law of motion in the vertical direction, we get: \(N - M*g = -M*g\) Solving for N: \(N = M*g - M*g = 0\)
02

Analyze Horizontal Forces

Now, let us analyze the forces acting horizontally on the body. The body is being pulled horizontally, and the frictional resistance opposing this pull is given by \(F_{friction} = \mu*N\), where \(\mu\) is the coefficient of friction.
03

Calculate Frictional Resistance

Now that we have the normal force N, we can calculate the frictional resistance. Substitute the value of N, which is 0, in the expression for F_friction: \(F_{friction} = \mu*N = \mu*0\)
04

Final Answer

From the above derivation, we find that the frictional resistance offered by the body is 0. Therefore, the answer is: (D) Zero

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Most popular questions from this chapter

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

For a particle rotating in a vertical circle with uniform speed, the maximum and minimum tension in the string are in the ratio \(5: 3\). If the radius of vertical circle is \(2 \mathrm{~m}\), the speed of revolving body is \(\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)\) (A) \(\sqrt{5} \mathrm{~m} / \mathrm{s}\) (B) \(4 \sqrt{5} \mathrm{~m} / \mathrm{s}\) (C) \(5 \mathrm{~m} / \mathrm{s}\) (D) \(10 \mathrm{~m} / \mathrm{s}\)

Two masses \(8 \mathrm{~kg}\) and \(12 \mathrm{~kg}\) are connected at the two ends of a light inextensible string that goes over a frictionless pulley. Find the tension the string when the masses are released. (A) \(96 \mathrm{~N}\) (B) \(80 \mathrm{~N}\) (C) \(100 \mathrm{~N}\) (D) None of these

A particle is moving along the circular path with a speed \(v\) and tangential acceleration is \(g\) at an instant. If the radius of the circular path be \(r\), then the net acceleration of the particle at that instant is (A) \(\frac{v^{2}}{r}+g\) (B) \(\frac{v^{2}}{r^{2}}+g^{2}\) (C) \(\left[\frac{v^{4}}{r^{2}}+g^{2}\right]^{\frac{1}{2}}\) (D) \(\left[\frac{v^{2}}{r}+g^{2}\right]^{\frac{1}{2}}\)

Reading of the spring scale in figure (B) (A) \(90 \mathrm{~N}\) (B) \(62.5 \mathrm{~N}\) (C) \(55 \mathrm{~N}\) (D) \(75 \mathrm{~N}\)
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