Question

A screw jack has square threads 50 mm mean diameter and 10 mm pitch. The load on the jack revolves with the screw. The coefficient of friction at the screw thread is 0.05. (1) Find the tangential force required at the end of 300 mm lever to lift a load of 6000 N. (2) State whether the jack is self-locking. If not, find the torque which must be applied to keep the load from descending.

Answer 1

Answer:

Concept:

Tangential force is defined as a force that always moves an object in the same direction. A body's motion is always perpendicular to the direction in which centrifugal force operates. It is not a tangential force as a result. The force operating on a moving body in the direction that is tangent to the body's curving path is referred to as tangential force. The acceleration will be negative if the object's velocity is positive. The force that pushes a body in a circular motion in the tangential direction of a curved path is known as tangential force. Let's respond, Finding tangential force. A tangential acceleration, that always originates from the axis of rotation and is perpendicular to the radius, leads to a tangential force.

Given:

Square threads on a screw jack have a mean diameter of 50 mm and a pitch of 10 mm. The jack's load spins around the screw. At the screw thread, the coefficient of friction is 0.05.

(1) To lift a 6000 N weight, calculate the tangential force needed at the end of the 300 mm lever.

(2) Indicate if the jack has a self-locking feature. If not, determine how much torque has to be provided to prevent the load from falling.

Find:

find the answers for the given questions

Answer:

$tan$θ $=\frac{p}{\pi d}$

       $=\frac{10}{\pi *50} \\=0.0637$

$tan$φ $=0.05$

φ $=2.8624$°

1). $P=\frac{d}{2R} *Wtan$(θ + φ)

       $=\frac{50}{2*300}$ $*6000$ $tan$(3.6426° + 2.8624°)

   $P = 57.01 N$

2) We have

$VR=\frac{2\pi R}{p} \\ =\frac{2\pi *300}{10} \\ =188.496$

$MA=\frac{W}{P} \\ = \frac{6000}{57.01} \\ =105.245$

$Efficiency=\frac{MA}{VR} \\\\ =\frac{105.245}{188.496} \\=0.5583\\\\Efficiency=55.83% > 50$% $> 50$

          The screw jack is indeed not self-locking as a result.  

the amount of torque necessary to prevent the weight from falling

$=\frac{d}{2} W tan$(θ − φ)

$=\frac{50}{2*600} *tan$ (3.6426° – 2.8624°)

$T = 204.3 N-mm$

Use the knowledge that F r = Rcos() and F t = Fsin() if an object is pinned at a point and a force F is applied at a distance R from the pin at an angle relative to a line to the centre.

Consider a mechanic applying a 20-newton force on the end of a wrench. She must exert the force from her current stance at a 120-degree angle with respect to the wrench.

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