A 140 m uniform ladder weighing 490 n rests against a...

A 14.0 m uniform ladder weighing 490 N rests against a frictionless wall. The ladder makes a 59.0° angle with the horizontal. (a) Find the horizontal and vertical forces (in N) the ground exerts on the base of the ladder when an 850-N firefighter has climbed 3.90 m along the ladder from the bottom. Horizontal force magnitude: N Direction: towards the wall/away from the wall Vertical force magnitude: N Direction: up/down (b) If the ladder is just on the verge of slipping when the firefighter is 9.40 m from the bottom, what is the coefficient of static friction between the ladder and the ground? (c) What If? If oil is spilled on the ground, causing the coefficient of static friction to drop to half the value found in part (b), what is the maximum distance (in m) the firefighter can climb along the ladder from the bottom before the ladder slips? Distance: m

Transcript

00:01 Hello students, in this question, a ladder is leaning against a wall making an angle 59 degree with a horizontal and the length of the ladder is given as 14 m and weight of the ladder is given as 490 n and we have to calculate the horizontal and vertical component of the force.

00:21 For this, for the torque calculating we have to take the summation of force is equal to 0 that is fx minus f wall is equal to 0 and vertical component is fy minus weight of the ladder and weight of the firefighter is acting along the opposite direction therefore minus which is equal to 0.

00:41 Let this be equation 1 and this be equation 2.

00:46 Now for calculating the force along the x that is x component of the force we have to write the equation vl into that is weight of the ladder into l by 2 sin theta minus weight of the firefighter into 3 .90 meter is the distance he has crossed on the ladder given in question and this is sin 59 degree theta is 59 degree here therefore theta may be 59 minus f wall this is the unknown quantity we have to find out into l cos theta equal to 0 therefore putting all the values we will be getting 490 n into l by 2.

01:32 L is 14 meter here therefore l by 2 is 7.

01:36 7 into sin 59 minus what is the weight of the firefighter? firefighter is 850 n into 3 .90 into sin 59 degree minus force against the wall into 14 cos 59 equal to 0.

01:59 So on plugging all this value we will be arriving at the equation minus 7 .211 into fire wall equal to minus 98 .57 therefore the force against wall is equal to 13 .68 newton here from the first equation we can come at fx is equal to f wall therefore this can be written as fx.

02:26 So the x component of the force is 13 .68 newton we have to calculate the y component therefore y component is equal to this is when it comes to the right hand side it is the addition of weight of the ladder that is 490 newton plus weight of the firefighter that is 850 newton which is equal to 1340 newton this is vertical component of the force.

02:55 So the horizontal and vertical component of the force is found out.

03:05 Now we have to calculate the coefficient of static friction when the firefighter climbs 3 .90 meter on the ladder.

03:21 So fx is equal to mu s dot fy.

03:26 So therefore mu s is equal to fx by fy what is fx here 13 .68 newton divided by 1340 newton so which is equal to 0 .0102.

03:43 So the static coefficient of static friction is equal to 0 .0102.

03:54 So now we have to calculate so now we have to calculate the distance that the firefighter can climb the distance the firefighter can climb safely when mu s nu, nu coefficient of static friction reduces to half of mu s...

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