00:01
Hi, in this question, the mass of the car, m is given as 1 ,130 kg.
00:11
The inclination of the ram, that is theta, theta is given as 25 degrees.
00:23
And the inclination of the cable 5.
00:28
This is the cable, this is making an inclination of 5 degrees with respect to the plane ramp 5 is the one as 31 degrees we need to draw the free body diagram then we need to determine the tension in the cable tension t in the cable then we need to answer how hard the surface of the ramp pushes the car see these are all the three parameters we need to determine this one is the reaction or the normal force exerted by the ramp on the car.
01:19
Let us go to the subdivision a.
01:23
The subdivision a, we need to draw the free body diagram.
01:28
This is the car.
01:29
The weight of the car will be acting in the vertically downward direction, that is mg.
01:37
And this weight will have two components.
01:39
One will be the perpendicular component, perpendicular to the ramp.
01:46
This will be mg.
01:47
Mg cos theta and another component will be parallel to the ramp.
01:56
This will be mg sine theta.
02:03
Then the rope is making an inclination of 5 degrees.
02:09
This one is theta.
02:13
This is the tension.
02:15
This is making an inclination with respect to the ramp to 5 degrees.
02:19
Therefore, this will have two components.
02:22
One will be the parallel component.
02:24
Another will be the perpendicular component.
02:27
This will be t cos 5 and this will be t sine 5.
02:39
Then the normal reaction will be exerted by the ram on the core.
02:45
This will be be n...