Question

A bag of cement of weight 300 N hangs from three ropes as shown in the figure below. Two of
the ropes make angles of 53.0° and 37.0° with the horizontal. If the system is in
equilibrium, find the tensions , , and in the ropes.

Answer 1

To find the tensions in the ropes, we can start by resolving the forces acting on the bag of cement in the vertical and horizontal directions.

Let's denote the tensions in the ropes as T1, T2, and T3. The weight of the bag of cement is 300 N acting downwards.

In the vertical direction:
T1cos53° + T2cos37° - 300N = 0 (as the system is in equilibrium)
T1cos53° = T2cos37° + 300N

In the horizontal direction:
T1sin53° = T2sin37°

Now we have two equations and two unknowns. We can solve for T1 and T2:

From the horizontal equation:
T1 = T2 * (sin37° / sin53°)

Substitute T1 into the vertical equation:
T2 * (sin37° / sin53°) * cos53° = T2cos37° + 300N
T2 * cos37° = T2cos37° + 300N
T2 = 300N / (cos37° - cos53°)

Now that we have found T2, we can find T1 using the equation for T1 in terms of T2:
T1 = T2 * (sin37° / sin53°)

Finally, we can find T3 by recognizing that the vertical component of T3 must balance the weight of the bag of cement:
T3cos53° = 300N
T3 = 300N / cos53°

Calculating the values of T1, T2, and T3 using the above equations will give you the tensions in the ropes.