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Q2 You are pulling on a block along a tilted road [on inclined with a force] P = 7N @ an angel B = 10° w.r.t vertical You know: M = 50kg ; $\theta$ = 25 and $\mu_s = 0.9$ $f_N = 5.73 - 482.6$ $f_N = 476.9$ $f_{fs} = \mu_s * f_N$ $f_{fs} = 0.9 * 476.9$ $f_{fs} = 429.1$ It will not move because $f_{fs}$ is greater than $f_{gx}$ $f_g = mg$ $f_g = 50 * 9.81$ $= 490.5N$ $f_N = 7 \cos 10 - 490.5$ $f_N = 483.6N$ $f_{fs} = \mu_s * f_N$ $f_{fs} = 0.9 * 483.6$ $f_{fs} = 435.2$ $f_g = 490.5$ $7 \sin 10 = 402N$ $7 \cos 10 = 5.73N$ $490.5 \cos 25 = 444.6$ $490.5 \sin 25 = 207.2$ $f_N = 490.5 \cos 25 - 7 \cos 10$ $f_N = 444.6 - 6.89 = 437.71$ $f_{fs} = 0.9 * 437.71 = 393.939$ $f_{gx} = 490.5 \sin 25 + 7 \sin 10$ $f_{gx} = 207.2 + 1.21 = 208.41$

          Q2
You are pulling on a block along a tilted road [on inclined with a force]
P = 7N @ an angel B = 10° w.r.t vertical
You know: M = 50kg ; $\theta$ = 25 and
$\mu_s = 0.9$

$f_N = 5.73 - 482.6$
$f_N = 476.9$

$f_{fs} = \mu_s * f_N$
$f_{fs} = 0.9 * 476.9$
$f_{fs} = 429.1$

It will not move because $f_{fs}$ is greater than
$f_{gx}$

$f_g = mg$
$f_g = 50 * 9.81$
$= 490.5N$
$f_N = 7 \cos 10 - 490.5$
$f_N = 483.6N$
$f_{fs} = \mu_s * f_N$
$f_{fs} = 0.9 * 483.6$
$f_{fs} = 435.2$

$f_g = 490.5$
$7 \sin 10 = 402N$
$7 \cos 10 = 5.73N$
$490.5 \cos 25 = 444.6$
$490.5 \sin 25 = 207.2$
$f_N = 490.5 \cos 25 - 7 \cos 10$
$f_N = 444.6 - 6.89 = 437.71$
$f_{fs} = 0.9 * 437.71 = 393.939$
$f_{gx} = 490.5 \sin 25 + 7 \sin 10$
$f_{gx} = 207.2 + 1.21 = 208.41$

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q2 you are pulling on a block along a tilted road on inclined with a force p 7n an angel b 10 wrt vertical you know m 50kg theta 25 and mus 09 fn 573 4826 fn 4769 ffs mus fn ffs 09 47 41798

Added by Kari J.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Vivek Singh

01/31/2022

Step 1

We are given the mass of the block, the angle of inclination of the road, the applied force, the angle of the applied force with respect to the vertical, and the coefficient of static friction. First, let's list the given values: Mass of the block, $M = 50 \text{ Show more…

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Q2 You are pulling on a block along a tilted road [on inclined with a force] P = 7N @ an angel B = 10° w.r.t vertical You know: M = 50kg ; $\theta$ = 25 and $\mu_s = 0.9$ $f_N = 5.73 - 482.6$ $f_N = 476.9$ $f_{fs} = \mu_s * f_N$ $f_{fs} = 0.9 * 476.9$ $f_{fs} = 429.1$ It will not move because $f_{fs}$ is greater than $f_{gx}$ $f_g = mg$ $f_g = 50 * 9.81$ $= 490.5N$ $f_N = 7 \cos 10 - 490.5$ $f_N = 483.6N$ $f_{fs} = \mu_s * f_N$ $f_{fs} = 0.9 * 483.6$ $f_{fs} = 435.2$ $f_g = 490.5$ $7 \sin 10 = 402N$ $7 \cos 10 = 5.73N$ $490.5 \cos 25 = 444.6$ $490.5 \sin 25 = 207.2$ $f_N = 490.5 \cos 25 - 7 \cos 10$ $f_N = 444.6 - 6.89 = 437.71$ $f_{fs} = 0.9 * 437.71 = 393.939$ $f_{gx} = 490.5 \sin 25 + 7 \sin 10$ $f_{gx} = 207.2 + 1.21 = 208.41$

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Transcript

00:01 Hello everyone, let us do the following question.

00:03 Here we have given two columns like this.

00:07 We have to tell which of the option is correct.

00:11 If we'll solve this equation, we know that mg of the block is acting vertically downward.

00:19 If we'll resolve this into two parts, this is theta and this is also theta, this part can be written as mg, cos theta.

00:27 And here the part can be written as mg mg sine theta and here there will be a normal reaction and as applied force is upward frictional force is acting downward frictional force is acting downward we have to solve the questions in the first if it is said that in the first option it is said that if force is equal to 30 newton and theta is equal to 30 degree then which of the option is correct.

01:04 If force is equal to 30 downward forces which are acting is mg sine theta.

01:12 Mg sine theta plus friction.

01:15 Here he will write value of this is mg sine theta.

01:20 Frictional force can be written as mu mg into.

01:25 Cost theta.

01:26 So here it will take m into g common.

01:30 Here it will take m into g common.

01:34 So here we will get this value as m g sine theta, mg sine theta plus mu into cost theta.

01:43 If we'll put the values here m is 10, g is 10, theta i sign 30 is 1 by 2, mu is 0 .2, 0 .8.

01:54 Here, cos 30 is can be written as under the root 3 by 2.

01:58 So, we will solve this value, but this value comes 5 .52 .384 newton.

02:08 But applied force is only 30 newton...

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