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Laws of Motion - Common Forces in Mechanics

Grade 11CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Weight (WW): It is the gravitational force with which the Earth pulls an object towards its center. It is always directed vertically downwards and is given by W=mgW = mg.

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Normal Reaction (NN): This is the component of the contact force perpendicular to the surface of contact. For a body resting on a horizontal plane, N=mgN = mg. On an inclined plane of angle θ\theta, N=mgcos⁡θN = mg \cos \theta.

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Tension (TT): A force exerted by a stretched string, rope, or chain on the objects attached to its ends. It always acts away from the body along the length of the string.

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Frictional Force (ff): A force that opposes the relative motion (or tendency of motion) between two surfaces in contact. It acts parallel to the surfaces. Static friction (fsf_s) adjusts itself to equal the applied force until the limiting friction (fs,maxf_{s,max}) is reached.

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Spring Force (FsF_s): When a spring is compressed or extended by a distance xx, it exerts a restoring force proportional to the displacement, expressed as Fs=−kxF_s = -kx, where kk is the spring constant.

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Centripetal Force (FcF_c): In circular motion, the net force directed towards the center required to keep a body of mass mm moving in a circle of radius rr at speed vv is Fc=mv2rF_c = \frac{mv^2}{r}.

📐Formulae

W=mgW = mg

fs,max=μsNf_{s,max} = \mu_s N

fk=μkNf_k = \mu_k N

Fs=−kxF_s = -kx

T=2m1m2m1+m2g (For a vertical pulley with two masses)T = \frac{2m_1 m_2}{m_1 + m_2} g \text{ (For a vertical pulley with two masses)}

a=m2−m1m1+m2g (Acceleration in an Atwood machine where m2>m1)a = \frac{m_2 - m_1}{m_1 + m_2} g \text{ (Acceleration in an Atwood machine where } m_2 > m_1)

💡Examples

Problem 1:

A block of mass m=10 kgm = 10 \text{ kg} is placed on a horizontal surface. A horizontal force of 100 N100 \text{ N} is applied to it. If the coefficient of static friction μs=0.4\mu_s = 0.4 and kinetic friction μk=0.3\mu_k = 0.3, calculate the net force acting on the block. (Take g=10 m/s2g = 10 \text{ m/s}^2)

Solution:

First, calculate the Normal force: N=mg=10×10=100 NN = mg = 10 \times 10 = 100 \text{ N}. Then, calculate limiting friction: fs,max=μsN=0.4×100=40 Nf_{s,max} = \mu_s N = 0.4 \times 100 = 40 \text{ N}. Since the applied force (100 N100 \text{ N}) is greater than fs,maxf_{s,max}, the block moves and kinetic friction acts. Kinetic friction fk=μkN=0.3×100=30 Nf_k = \mu_k N = 0.3 \times 100 = 30 \text{ N}. Net force Fnet=Fapplied−fkF_{net} = F_{applied} - f_k: 100−3070\begin{array}{r} 100 \\ - 30 \\ \hline 70 \end{array} Fnet=70 NF_{net} = 70 \text{ N}.

Explanation:

To determine the net force, we first check if the applied force exceeds the static friction threshold. Since it does, we subtract the kinetic friction from the applied force to find the resultant force causing acceleration.

Problem 2:

A spring with a spring constant k=500 N/mk = 500 \text{ N/m} is compressed by x=0.2 mx = 0.2 \text{ m}. Calculate the magnitude of the restoring force exerted by the spring.

Solution:

Using Hooke's Law: F=kxF = kx. F=500×0.2F = 500 \times 0.2 500×0.2100\begin{array}{r} 500 \\ \times 0.2 \\ \hline 100 \end{array} F=100 NF = 100 \text{ N}.

Explanation:

The magnitude of the spring force is directly proportional to the displacement from the equilibrium position, as defined by F=kxF = kx.