Question

The systems shown below are in equilibrium (with m = 1.80 kg and ? = 25.0�). If the spring scales are calibrated in newtons, what do they read? Ignore the masses of the pulleys and strings and assume the pulleys and the incline are frictionless. scale in (a) N scale in (b) N scale in (c) N scale in (d) N

The systems shown below are in equilibrium (with m = 1.80 kg and θ = 25.0°). If the spring scales are calibrated in newtons, what do they read? Ignore the masses of the pulleys and strings and assume the pulleys and the incline are frictionless.

scale in (a)	N
scale in (b)	N
scale in (c)	N
scale in (d)	N

Answer 1

Concepts and reason

The main concepts be used to solve the given problem are equilibrium of force and Newton’s second law of motion.

Initially, the force applied on the spring scale in each case will be calculated by the applying the equilibrium condition for force. Later, the force applied on the spring scale in each case will be dependent on the weight of the ball or blocks. So, according to the equilibrium of force, the net force must be equated to zero. Finally, In the given scenario, the weight is supported by tension force. This tension force is measured by the scale.

Fundamentals

Newton’s Second Law: The Newton’s second law states that the net force on an object is the product of mass of the object and final acceleration of the object.

The equation of the Newton’s second law is,

Here, is the net force on the object, is mass of the object, and is the acceleration of the object.

The equilibrium condition for the force gives,

Here, is force.

Weight of an object is the force with which the earth pulls an object towards its center.

The formula for weight is,

Here, is the mass of object and is the acceleration due to gravity.

(a)

The force applied by both the balls on the spring scale will be identical since the weight of both the balls is same.

Apply the Newton’s second law to any of the block. The weight of the block is acting in downward direction and the tension is acting in upward direction. So,

$EF=T-mg $

Use equilibrium condition,

Substitute for and for in the above equation.

(b)

The force applied by both the balls on the spring scale will be identical since the weight of both the balls is same.

Apply the Newton’s second law to any of the block. The weight of the block is acting in downward direction and the tension is acting in upward direction. So,

Use equilibrium condition,

Substitute for and for in the above equation.

[Hint for next step]

Apply Newton’s second law to any of the block in figure (c) and equate that equation to zero and finally, solve the resulting equation for the tension force (T) i.e. the force measured by the scale.

(c)

The force applied on the spring scale will be balanced by the weight of both the blocks.

Apply the Newton’s second law to the given system,

Use equilibrium condition,

Substitute for and for in the above equation.

(d)

The figure given below shows the free body diagram of the block on the inclined plane.

In the above figure is the mass of the block, is acceleration due to gravity and is the angle of inclination of the inclined plane.

Split the components of the weight of the block along the x and y axis.

The weight component along the x axis is,

The weight component along the y axis is,

Apply the Newton’s second law along the incline.

Use the equilibrium condition,

Substitute for, for and for.

Ans: Part a

The reading on the spring scale will be.

Part b

The reading on the spring scale will be.

Part c

The reading on the spring scale will be.

Part d

The reading on the spring scale will be.