Your videos have been a great help, but this one is pretty convoluted. I approached this problem as a Three-Force Member. The lines of action from the weight and normal force at A intersect at a point. Therefore, the line of action of force B must also intersect at this point. We make some triangles to find the x and y distances to determine the angle of the force then use the weight calculation and the angle we found to determine the rest.
why didnt we use the (summation) Fx=0 equation to get the x component of the unknown force P or use the y component of the same force to determine P when we know the angle it makes with the vertical, i.e. 30 degrees ?
With all due respect to the video creator, this explanation is incorrect. Part of your answer is finding the angle of the forces at B, so you cannot assume it is 60. I am not sure if his processes are incorrect other than that though. The way you find the force at A is by using the moment equation at point B. The moment will be 0 and you use the x and y components of the weight and A force to set up an equation which allows you to find A.
U are legend but some hints Take 60 degree horizontal line straight up where .2m is now .2m is 90 degree with line adjacent to 60 degree mean right triangle soo infront of 60 deg is opposite side other is adjacent Find adjacent Subtract from .4 .4 is straight line without bend given on other top corner Soo .2 & subtracted lenth are x and y components of force Find angle opp/adj about 54.46 Now you have force components Bxcos & bysin Now come down Don't forget add .1 when determine vertically distance to wieght this side angle is 60 other side it will become 30 soo .4cose30 Is lenth from out side the tyre to vertically w add .1 which is inside tyre lenth from centre Now Force bx is .4+.5.+.4 far away By is .2 minus .1÷cos30
Your videos have been a great help, but this one is pretty convoluted. I approached this problem as a Three-Force Member. The lines of action from the weight and normal force at A intersect at a point. Therefore, the line of action of force B must also intersect at this point. We make some triangles to find the x and y distances to determine the angle of the force then use the weight calculation and the angle we found to determine the rest.
I want to know why there is not vertical component at the wheel
can you give me an explanation of why did let the force exerted on the grip as parallel to length of the box
why didnt we use the (summation) Fx=0 equation to get the x component of the unknown force P or use the y component of the same force to determine P when we know the angle it makes with the vertical, i.e. 30 degrees ?
With all due respect to the video creator, this explanation is incorrect. Part of your answer is finding the angle of the forces at B, so you cannot assume it is 60. I am not sure if his processes are incorrect other than that though. The way you find the force at A is by using the moment equation at point B. The moment will be 0 and you use the x and y components of the weight and A force to set up an equation which allows you to find A.
The answers in the text book are
P = 600 N, theta = 48 degrees CW from x axis, N_A = 442 N
Yeah, someone called him out and it seems he reads answers off Chegg
I second that... you're right
I don’t get where you got that the force is 60 degrees off the X axis
very useful thank you
U are legend but some hints
Take 60 degree horizontal line straight up where .2m is now .2m is 90 degree with line adjacent to 60 degree mean right triangle soo infront of 60 deg is opposite side other is adjacent
Find adjacent
Subtract from .4
.4 is straight line without bend given on other top corner
Soo .2 & subtracted lenth are x and y components of force
Find angle opp/adj about 54.46
Now you have force components
Bxcos & bysin
Now come down Don't forget add .1 when determine vertically distance to wieght this side angle is 60 other side it will become 30 soo .4cose30
Is lenth from out side the tyre to vertically w add .1 which is inside tyre lenth from centre
Now
Force bx is .4+.5.+.4 far away
By is .2 minus .1÷cos30
This guy just reads answers off Chegg.
A free body diagram would've helped