Question

In the figure, if the tension in string 1 is 34 N (40 degrees to the left) and the tension in string 2 is 24 N ( to the right) what is the mass of the object shown?

Answer 1

Answer:

3.7 kg

Explanation:

Start off by drawing and labeling a free-body diagram for the object. I have attached an image that I drew that shows all of the relevant forces acting on the object.

On the free-body diagram, we have the tension forces (T₁ and T₂), their x and y components, and the weight force (w).

We want to start off by finding the angle theta (Θ) so that we can use this angle when using Newton's Second Law later on in the problem to calculate the mass of the object.

We can find the angle Θ by using the fact that there are no unbalanced horizontal forces acting on the object, so the net force is zero.

This means that T₁ sin(40) is equal to T₂ sin(Θ). Set them equal to each other in an equation and solve for the angle Θ. We are given the tension forces: T₁ = 34 N and T₂ = 24 N.

• T₁ sin(40) = T₂ sin(Θ)
• 34 * sin(40) = 24 * sin(Θ)

Evaluate 34 * sin(40) and divide it by 24 to isolate sin(Θ).

• 0.9106157804 = sin(Θ)

Take the inverse sine of both sides of the equation.

• Θ = sin⁻¹(0.9106157804)
• Θ = 65.59058699°

Now that we have the angle Θ, we can use this to set up an equation involving mass using Newton's Second Law.

We know that there is a weight force directed downwards on the object, and this weight force is equal to mg (mass · gravitational acceleration).

There are two forces acting upwards on the object: the x-components of the two tension forces.

Using the fact that the object is not moving vertically, we know that the net force must equal 0 (since F = ma and acceleration = 0 m/s², therefore F = 0 N).

The sum of the forces in the y-direction will be equal to 0. Setting the upwards direction to be positive and the downwards direction to be negative, we can write this equation:

• ∑Fᵧ =  T₁ cos(40) + T₂ cos(65.59058699) - mg = 0

Add mg (weight force) to both sides of the equation.

• T₁ cos(40) + T₂ cos(65.59058699) = mg

Substitute 34 for T₁ and 24 for T₂.

• 34 * cos(40) + 24 * cos(65.59058699) = mg

The gravitational acceleration on Earth is 9.8 m/s². Substitute this value into the equation and solve for m, the mass of the object.

• 34 * cos(40) + 24 * cos(65.59058699) = m(9.8)

Evaluate the left side of the equation.

• 35.96360799 = 9.8m

Divide both sides of the equation by 9.8.

• m = 3.669755918

Round this to two significant figures since the problem gives us the tension forces to two sig figs.

• m ≈ 3.7 kg

The mass of the object shown is around 3.7 kg.

Answer 2

Answer:

3.7 kg

Explanation:

Start off by drawing and labeling a free-body diagram for the object. I have attached an image that I drew that shows all of the relevant forces acting on the object.

On the free-body diagram, we have the tension forces (T₁ and T₂), their x and y components, and the weight force (w).

We want to start off by finding the angle theta (Θ) so that we can use this angle when using Newton's Second Law later on in the problem to calculate the mass of the object.

We can find the angle Θ by using the fact that there are no unbalanced horizontal forces acting on the object, so the net force is zero.

This means that T₁ sin(40) is equal to T₂ sin(Θ). Set them equal to each other in an equation and solve for the angle Θ. We are given the tension forces: T₁ = 34 N and T₂ = 24 N.

T₁ sin(40) = T₂ sin(Θ)

34 * sin(40) = 24 * sin(Θ)

Evaluate 34 * sin(40) and divide it by 24 to isolate sin(Θ).

0.9106157804 = sin(Θ)

Take the inverse sine of both sides of the equation.

Θ = sin⁻¹(0.9106157804)

Θ = 65.59058699°

Now that we have the angle Θ, we can use this to set up an equation involving mass using Newton's Second Law.

We know that there is a weight force directed downwards on the object, and this weight force is equal to mg (mass · gravitational acceleration).

There are two forces acting upwards on the object: the x-components of the two tension forces.

Using the fact that the object is not moving vertically, we know that the net force must equal 0 (since F = ma and acceleration = 0 m/s², therefore F = 0 N).

The sum of the forces in the y-direction will be equal to 0. Setting the upwards direction to be positive and the downwards direction to be negative, we can write this equation:

∑Fᵧ =  T₁ cos(40) + T₂ cos(65.59058699) - mg = 0

Add mg (weight force) to both sides of the equation.

T₁ cos(40) + T₂ cos(65.59058699) = mg

Substitute 34 for T₁ and 24 for T₂.

34 * cos(40) + 24 * cos(65.59058699) = mg

The gravitational acceleration on Earth is 9.8 m/s². Substitute this value into the equation and solve for m, the mass of the object.

34 * cos(40) + 24 * cos(65.59058699) = m(9.8)

Evaluate the left side of the equation.

35.96360799 = 9.8m

Divide both sides of the equation by 9.8.

m = 3.669755918

Round this to two significant figures since the problem gives us the tension forces to two sig figs.

m ≈ 3.7 kg

The mass of the object shown is around 3.7 kg.