Consider an in-extensible string passing over an ideal pulley connected to a block of mass $m$ as shown:

The ends are pulled with a force $F$

Am I correct in saying that the block will experience a reaction force $\sqrt{2}F$? The forces acting are perpendicular to each other and the support of the pulley is inclined at an angle of $45^\circ$ with horizontal.

In other words can the forces acting on the string be vectorially added to find the resultant reaction force exerted by pulley?

  • 1
    $\begingroup$ I encountered this specific scenario while solving an entirely different question. That is the reason I havent given any context. Im not sure about why this is tagged as a homework question. This is a purely conceptual doubt $\endgroup$ – newbie105 Jan 2 at 16:27

Yes, this is correct. However, the pulley will also experience a reaction force from the connection to the block.

Two things to note:

  1. Since this is a pulley, there is also torque, since the forces are not acting on the center of mass (in this case the torques cancel).
  2. As soon as the pulley starts moving, your analysis will change, because the force change direction.
  • $\begingroup$ Can you explain why the directions of the applied forces change? Wouldn't they be acting horizontally and vertically just like before? $\endgroup$ – newbie105 Jan 2 at 16:49
  • $\begingroup$ @newbie105 It depends on how the ropes are attached. If the ropes are fixed at the point, the rope will have a different angle, and therefore also the force. $\endgroup$ – Bernhard Jan 2 at 18:18
  • $\begingroup$ Can the downvoter explain their vote? $\endgroup$ – Bernhard Jan 3 at 8:47
  • 1
    $\begingroup$ @Buraian I would consider it a normal force, not a friction force. $\endgroup$ – Bernhard Jan 3 at 8:48
  • $\begingroup$ Your answer would be complete if you also address the more general question of the possibility of vector addition for any angle. I would also like to have a proof for the same. $\endgroup$ – newbie105 Jan 3 at 12:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.