Jason is pulling a box across the room. He is pulling with a force of 19 newtons and his arm is making a 30 angle with the horizontal, what is the vertical component of the force he is pulling with?

Jason is pulling a box across the room. He is pulling with a force of 14 newtons and his arm is making a 79 angle with the horizontal, what is the horizontal component of the force he is pulling with?

Jason is pulling a box across the room. He is pulling with a force of 23 newtons and his arm is making a 30 angle with the horizontal, if the box weighs 24 newtons what is the net force on the box in the vertical direction? Treat up as the positive direction, and down as the negative direction.

One force of magnitude 68 acts on an object. Another force of magnitude 24 acts on the object at a right angle, what is the magnitude of the resultant force on the object?

I need help badly. Can yall please try and answer as many as you can? I would really appreciate it!

I would suggest using it on a calculator as it can become very time consuming on paper. Sine (angle) times force. Sin(30) times 19. I believe the answer is 9.5.

Question 2:

You would use Sine for vertical directions. I suggest doing it by calculator.

Sin(30) times 23 which would give you 11.5. sin(30) x 23. Then, you would subtract the weight of the box from it. so, 11.5 - 24. The net force is -12.5.

Question 3:

Use pythagoreans theorem √ x²+y² = ||V|| or a²+b² = c²

√68²+24² = √c²

c = 72.111 or 72.1 rounded.

adioabiola adioabiola · Answer 2 · 2021-09-15T02:36:00+02:00

We are to solve for the vertical and horizontal component of the force he is pulling with.

1. The vertical component of Jason's force is 9.5Newtons.

2.The horizontal component of Jason's force is 2.67Newtons.

3. Therefore, the net force in the vertical direction is -12.5Newtons.

4.Therefore, Resultant, R is 72.11

Question 1:

If Jason is pulling with a force of 19 Newtons and his arm is at an angle of 30° with the horizontal.

By resolving Jason's force of pull in the vertical direction, Fy = 19 × Sin 30° = 9.5 Newtons.

The vertical component of Jason's force is 9.5 Newtons.

Question 2:

Also,If Jason is pulling with a force of 14 Newtons and his arm is at an angle of 79° with the horizontal.

By resolving Jason's force of pull in the horizontal direction, Fx = 14 × Cos79° = 2.67 Newtons.

The horizontal component of Jason's force is 2.67Newtons.

Question 3:

If Jason is pulling with a force of 23Newtons and his arm is at an angle of 30° with the horizontal.

By resolving Jason's force of pull in the vertical direction, Fy = 23 × Sin 30° = 11.5Newtons.

The vertical component of Jason's force of pull is, 11.5Newtons.

If the box weighs 24Newtons.

By treating up as the +ve vertical direction and down as the -ve vertical direction,

Therefore, the weight of the box acts in the -ve vertical direction, while Jason's vertical force component acts in the +ve vertical direction.

Therefore, the net force in the vertical direction is = 11.5Newtons + (-24Newtons)

Therefore, the net force in the vertical direction is -12.5Newtons.

Question 4:

If there are two forces at right angles to eachother, one of magnitude, 68 and the other of magnitude, 24.

The resultant force on the object can be obtained by Pythagoras theorem (Triangle law of forces).

Resultant, R = √(68²+24²) = √5200.

Therefore, Resultant, R = 72.11