/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 154 A block rests on a rough incline... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 154

A block rests on a rough inclined plane making an angle of \(30^{\circ}\) with the horizontal. The co-efficient of static friction between the block and the plane is \(0.8\). If the frictional force on the block is \(10 \mathrm{~N}\), the mass of the block (in \(\mathrm{kg}\) ) is (Taking \(g=10 \mathrm{~m} / \mathrm{s}^{2}\) ) (A) \(2.0\) (B) \(4.0\) (C) \(1.6\) (D) \(2.5\)

Short Answer

Expert verified
The mass of the block is approximately \(1.6 \mathrm{~kg}\) (C).

Step by step solution

01

Draw a force diagram

Draw a force diagram for the block, which includes the weight of the block (mg), normal force (N) perpendicular to the inclined plane, and frictional force (f) opposing the block's motion.
02

Calculate the frictional force

The frictional force (f) can be calculated using the formula: \(f = \mu N\) where \(\mu\) is the coefficient of static friction and \(N\) is the normal force.
03

Calculate the normal force

To find the normal force (N), we need to break down the weight of the block into two components: component parallel to the incline (\(mg\sin{\theta}\)), and component perpendicular to the incline (\(mg\cos{\theta}\)). The normal force is equal to the component of the weight perpendicular to the incline: \(N = mg\cos{\theta}\)
04

Substitute the values

Now substitute the given values in the equation: \(f = \mu N\) \(10 = 0.8 \times (m)(10\cos{30^{\circ}})\)
05

Solve for the mass (m)

Now we can solve for the mass of the block (m): \(10 = 0.8 \times (m)(10\times(\frac{\sqrt{3}}{2}))\)
06

Simplify and select the correct answer

Simplify the equation and solve for m: \(10 = 4\sqrt{3} \times m\) \(m = \frac{10}{4\sqrt{3}}\) \(m = \frac{5}{2\sqrt{3}}\) \(m = \frac{5\sqrt{3}}{6}\) The closest answer from the options is \(1.6 \mathrm{~kg}\) (C).

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Most popular questions from this chapter

An insect crawls up a hemispherical surface very slowly (Fig. \(3.88\) ). The coefficient of friction between the insect and the surface is \(1 / 3 .\) If the line joining the centre of hemispherical surface to the insect makes an angle \(\alpha\) with the vertical, the maximum possible value of \(\alpha\) is given by (A) \(\cot \alpha=3\) (B) \(\tan \alpha=3\) (C) \(\sec \alpha=3\) (D) \(\operatorname{cosec} \alpha=3\)

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

Three blocks \(m_{1}, m_{2}\) and \(m_{3}\) of masses \(8 \mathrm{~kg}, 3 \mathrm{~kg}\) and \(1 \mathrm{~kg}\) are placed in contact on a smooth surface. Forces \(F_{1}=140 \mathrm{~N}\) and \(F_{2}=20 \mathrm{~N}\) are acting on blocks \(m_{1}\) and \(m_{3}\), respectively, as shown. The reaction between blocks \(m_{2}\) and \(m_{3}\) is (A) \(2.5 \mathrm{~N}\) (B) \(7.5 \mathrm{~N}\) (C) \(22.5 \mathrm{~N}\) (D) \(30 \mathrm{~N}\)

Consider the system shown in Fig. 3.78. The wall is smooth, but the surface of blocks \(A\) and \(B\) in contact are rough. The friction on \(B\) due to \(A\) in equilibrium is (A) Upward (B) Downward (C) Zero (D) The system cannot remain in equilibrium.

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 \(M / 2 \mathrm{~kg}\)
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