#### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-5470-145

+91 7353221155

CART 0

• 0
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

# Find acceleration of masses m1 and m2 in the set up shown, pulleys and string are massless. Neglect any friction.

Shane Macguire
30 Points
8 years ago

First we notice that the acceleration a1 =  a2. This can be seen as follows. Suppose m2 descends by Dx. Then the reduction in the length of string between the point of attachment on the wall and pulley attached to the end of incline also decrease by Dx. The pulley attached to m1 then moves by Δx/2.

If T is the tension in the string then the free body diagram for m1 is as shown in figure.

Shane Macguire
30 Points
8 years ago

First we notice that the acceleration a1 = 1/2 a2. This can be seen as follows. Suppose m2 descends by Δx. Then the reduction in the length of string between the point of attachment on the wall and pulley attached to the end of incline also decrease by Δx. The pulley attached to m1 then moves by Δx/2.

If T is the tension in the string then the free body diagram for m1 is as shown in figure.

x component : 2T – m1gsinθ = m1a1                               (I)

y component : N – m1gcosθ = 0

The free body diagram for m2 is                                    (II)

m2g – T = m2a2                                                           (III)

Solving equations (I) and (II)