00:01
A horse is harnessed to a sled having a mass of 236 kilogram including supplies.
00:06
So a horse must exert a force of force which is exceeding 1 to 4 .0 newton at an angle of 35 degree in order to get the sled moving.
00:16
So the situation is like this.
00:19
There is a mass and there is a force applied at an angle of 35 degrees by the horse.
00:25
And this force must be greater than 1 to 4 .40 newton in order to get the sled.
00:30
Moving so can you get the normal force on this slide when the magnitude of the applied force is 1 240 1 240 so there is definitely a static friction on it so the normal reaction x over here the weight is over here the friction is over here and if you're going to make the components of this 1 240 that's going to be horizontal component is 1 240 cost 35 degrees and the vertical component is going to be 1 240 or let's draw it over here 1 240 sign 35 degrees so these are the two components of the forces we are talking about particle okay so since uh the 1 2 the the horse exerts the force exceeding 1 240 will make it moving so at 1 to 4 0 it is actually static so so if we equate the vertical forces, then we have r plus 1240 sine 35 degrees is equal to the weight.
01:42
And weight is mg and the mass is 236 kilograms.
01:47
So from this, the value of r comes out as, let me grab my calculator here, that's 236 times 9 .8 minus 1240 times sine 35.
01:58
So this is coming as 1601 .57 newton.
02:04
So this is the reaction force or the normal force.
02:11
Then in the next part, they are talking about when the coefficient of static friction between the sled in the ground.
02:19
So for that we equate the horizontal forces.
02:22
So for that the friction is going to be equal to 1 -240 cost 35...