StudySmarter AI is coming soon!

- :00Days
- :00Hours
- :00Mins
- 00Seconds

A new era for learning is coming soonSign up for free

Suggested languages for you:

Americas

Europe

Q17P

Expert-verifiedFound in: Page 381

Book edition
9th Edition

Author(s)
Raymond A. Serway, John W. Jewett

Pages
1624 pages

ISBN
9781133947271

**Figure P12.17 shows a claw hammer being used to pull a nail out of a horizontal board. The mass of the hammer is ${\mathbf{1}}{\mathbf{.}}{\mathbf{00}}{\mathit{k}}{\mathit{g}}$. A force of ${\mathbf{150}}{\mathit{N}}$is exerted horizontally as shown, and the nail does not yet move relative to the board. Find (a) the force exerted by the hammer claws on the nail **

(a) The force exerted by the hammer claws on the nail will be $1.04\text{kN at 60}\xb0$ upward and to the right.

Mass of the hammer $M=1.00Kg$

Force exerted horizontally = $\text{150N}$

**Force is physical influence on any material or object that can change the magnitude and direction of the motion of any object. It can be mainly denoted by **F.

** **

By the given figure if we will take the single point of contact as *P* then the exerted force by nail on the hammer claws will be *R*. The mass of hammer is given that is *M* = 1.00Kg and the normal force that will be exerted on the hammer at point *P* be *n* and the friction acting on it is *f*.

Then by the moments P,

$\left(R\mathrm{sin}30\xb0\right)0+\left(R\mathrm{sin}30\xb0\right)\left(5.00\text{cm}\right)+Mg\left(0\right)-\left(\text{150N}\right)\left(30.0\text{cm}\right)=0\phantom{\rule{0ex}{0ex}}R=1039.2N\phantom{\rule{0ex}{0ex}}R=1.04\text{kN}\phantom{\rule{0ex}{0ex}}$

The Force exerted by the hammer on the nail is equal opposite in direction but equal in the magnitude, so it will be $1.04\text{kN at 60}\xb0$ upward and to the right.

94% of StudySmarter users get better grades.

Sign up for free