In the arrangement shown in the figure [sin 37°=3/5]

Question

In the arrangement shown in the figure [sin 37°=3/5] <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/5c4ccb3cb377dac5dccaa9caa22307fd-.png" /> (A) direction of force of friction is up the plane (B) the magnitude of force of friction is zero (C) the tension in the string is 20 N (D) magnitude of force of friction is 56 N

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/018c34a6ecd4969129e88eaf4d059579-.png" /> Net driving force = 60 - 40 = 20 N (down the plane) As resisting force is greater than net driving force, the friction will be static of nature and friction force is 20 N (up the plane)

Q. In the arrangement shown in the figure $[s in 3 7^{\circ} = 3/5]$

direction of force of friction is up the plane

the magnitude of force of friction is zero

the tension in the string is $20 N$

magnitude of force of friction is $56 N$

Net driving force $= 60 - 40 = 20 N$ (down the plane)

As resisting force is greater than net driving force, the friction will be static of nature and friction force is $20 N$ (up the plane)

Q. In the arrangement shown in the figure [sin37∘=3/5]

direction of force of friction is up the plane

the magnitude of force of friction is zero

the tension in the string is 20N

magnitude of force of friction is 56N

Net driving force =60−40=20N (down the plane) As resisting force is greater than net driving force, the friction will be static of nature and friction force is 20N (up the plane)

Q. In the arrangement shown in the figure $[s in 3 7^{\circ} = 3/5]$

the tension in the string is $20 N$

magnitude of force of friction is $56 N$

Net driving force $= 60 - 40 = 20 N$ (down the plane)

As resisting force is greater than net driving force, the friction will be static of nature and friction force is $20 N$ (up the plane)