I can not calculate these Newton equations

[Apprentice123]

A block B of 20 kg is trapped by a wire of 2 m a car A of 30 kg. Determine (a) the acceleration of the car (b) the tension of the wire, immediately after the system was abandoned the rest, in the position shown in the figure. Despise is the friction.

Answer:
(a) 6,17 m/s^2
(b) 144 N

My attempt
CAR

ZFx = mA . a
Tcos(65) = 30a

Block B
ZFx = mB . a
Tsin(65) = mB.a

I can not calculate these equations

 

[Apprentice123]

A block B of 20 kg is trapped by a wire of 2 m a car A of 30 kg. Determine (a) the acceleration of the car (b) the tension of the wire, immediately after the system was abandoned the rest, in the position shown in the figure. Despise is the friction.

Answer:
(a) 6,17 m/s^2
(b) 144 N

My attempt
CAR

ZFx = mA . a
Tcos(65) = 30a

Block B
ZFx = mB . a
Tsin(65) = mB.a

I can not calculate these equations

The equations are correct?

 

[drizzle]

why are you writing ZFx? do you mean the x component of F? but the car moves in a direction where it has x and y components of the force [Fx and Fy]
for the tension of the wire there’s no Fx

 

[Apprentice123]

why are you writing ZFx? do you mean the x component of F? but the car moves in a direction where it has x and y components of the force [Fx and Fy]
for the tension of the wire there’s no Fx

ZF => Sum of forces

 

[PhanthomJay]

It is probably best to choose the x-axis as the axis parallel to the incline, both when looking at the forces on block A and the forces on Block B. If you look at block B first and sum forces in the "x" direction, that is, in the direction parallel to the incline, then it's Tsin25[/color] plus the weight force component =20a, where 'a' is the acceleration in that x direction parallel to the incline. Then look at Block A using the same set of axes, and don't forget to include the weight (gravity) force acting down the incline.

 

[Apprentice123]

It is probably best to choose the x-axis as the axis parallel to the incline, both when looking at the forces on block A and the forces on Block B. If you look at block B first and sum forces in the "x" direction, that is, in the direction parallel to the incline, then it's Tsin25[/color] plus the weight force component =20a, where 'a' is the acceleration in that x direction parallel to the incline. Then look at Block A using the same set of axes, and don't forget to include the weight (gravity) force acting down the incline.

Not Tcos (25) ?
The acceleration of the block is the same car ?

 

[PhanthomJay]

Sorry, I may have misread this problem. I believe the problem is asking for the car acceleration and the tension in the wire holding the 20 kg mass (Block B) at the instant the car is released from rest, while the wire is in the vertical position. If that is the case, it seems you should look at block B and sum forces in the vertical y direction to get 20g - T = 20a_y. Then look at the car, and sum forces along the incline ( the weight and tension components acting along the incline as the assumed x direction) equal to 30a_x, where a_x and a_y are related by a_x =a_y/sin25.

 

[Apprentice123]

Sorry, I may have misread this problem. I believe the problem is asking for the car acceleration and the tension in the wire holding the 20 kg mass (Block B) at the instant the car is released from rest, while the wire is in the vertical position. If that is the case, it seems you should look at block B and sum forces in the vertical y direction to get 20g - T = 20a_y. Then look at the car, and sum forces along the incline ( the weight and tension components acting along the incline as the assumed x direction) equal to 30a_x, where a_x and a_y are related by a_x =a_y/sin25.

How to calculate sum of forces in the X axis? Only has the weight

 

[PhanthomJay]

How to calculate sum of forces in the X axis? Only has the weight

For block B, I took the y-axis as vertical and the x-axis as horizontal, and summed forces in the y direction. But for the free body diagram of car A, I chose the x-axis to be the axis parallel to the incline, that is, 25 degrees up from the horizontal. So acting along the x-axis , you have the component of the cars weight (mgsin 25) and also the component of the Tension force (Tsin25). Their sum is 30a_x. You will need to solve 2 equations with 2 unknowns, noting first the relation beteen a_x and a_y (since the block and car are connected, they must accelerate at the same rate).

 

[Apprentice123]

For block B, I took the y-axis as vertical and the x-axis as horizontal, and summed forces in the y direction. But for the free body diagram of car A, I chose the x-axis to be the axis parallel to the incline, that is, 25 degrees up from the horizontal. So acting along the x-axis , you have the component of the cars weight (mgsin 25) and also the component of the Tension force (Tsin25). Their sum is 30a_x. You will need to solve 2 equations with 2 unknowns, noting first the relation beteen a_x and a_y (since the block and car are connected, they must accelerate at the same rate).

Thanks

 

[PhanthomJay]

Clarificaton Edit: For block B, I took the y-axis as vertical and the x-axis as horizontal, and summed forces in the y direction. But for the free body diagram of car A, I chose the x-axis to be the axis parallel to the incline, that is, 25 degrees up from the horizontal. So acting along the inclined x-axis , you have the component of the cars weight (mgsin 25) and also the component of the Tension force (Tsin25). Their sum is 30a_x. You will need to solve 2 equations with 2 unknowns, noting first the relation beteen a_x and a_y (since the block and car are connected, they must initially vertically accelerate at the same rate a_y, but not so horizontally...until ultimately after a very brief time, once the lower block swings into its final position with the rope perpendicular to the incline, they accelerate together in both the x and y directions, at the same rate, but different from the initial value of the calcualted accelerataion.